STRUCTURAL  ENGINEERING 


By 
J.   E,   KIRKHAM 

Professor  of  Structural  Engineering,  Iowa  State  College.     Consulting 

Bridge  Engineer,   Iowa  Highway  Commission.       Formerly 

Designing  Engineer  with  American  Bridge  Co. 


FIRST  EDITION 
THIRD  IMPRESSION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
NEW  YORK:  370  SEVENTH  AVENUE 

LONDON :  6  &  8  BOUVERIE  ST.,  E.  C,  4 

1914 


COPYRIGHT  1914 

BY 

THE  MYRON  C.  CLARK  PUBLISHING  CO. 
CHICAGO 


TABLE  OF  CONTENTS 


CHAPTER   I. 
Preliminary. 

PAGES 

Definitions,  structural  material,  riveting,  method  of  procedure  in  designing,  and 

fabrication    1—8 


CHAPTER  II. 
Structural  Draughting1. 

Equipment,  free-hand  lettering,   drawings,  and   hints  regarding  shop  drawings 

and  bills 9-13 


CHAPTER  III. 
Fundamental  Elements  of  Structural  Mechanics. 

Definitions,  forces,  elasticity,  distortion,  equilibrium  polygon,  center  of  gravity, 

moment  of  inertia,  radius  of  gyration 14-50 


CHAPTER   IV. 
Theoretical  Treatment  of  Beams. 

Shearing  stresses,  bending  stresses,  reactions  and  deflections  of  cantilever,  simple, 
over-hanging,  fixed  at  one  end  and  supported  at  other,  fixed  and  continuous 
beams 51-105 


CHAPTER  V. 
Theoretical  Treatment  of  Columns. 

Rankine's  formula,  straight  line  formula,  eccentrically  loaded  columns,  and  ex- 
amples      106-114 


CHAPTER  VI. 
Rivets,  Fins,  Rollers  and  Shafting*. 

Bearing,  shearing  and  bending  stresses  on  rivets  and  pins,  allowable  pressure 

on  rollers  and  stresses  on  shafting   115-123 


CHAPTER  VII. 

Maximum  Reactions,  Shears,  and  Bending  Moments  on  Beams  and  Trusses  and 

Stresses  in  Trusses. 

Maximum  reactions,  shears  and  bending  moments  on  simple  beams  and  trusses 
due  to  both  uniform  and  concentrated  loads,  and  stresses  in  simple  trusses 
due  .to  panel  loads  124-143 

iii 


jv  CONTENTS 

CHAPTER  VIII. 
Graphic  Statics. 

PAGES 

Definition  and  limitation,  graphical  extermination  of  reactions  and  bending 
moments  on  simple  beams,  graphical  determination  of  stresses  in  trusses 
and  drawing  of  an  equilibrium  polygon  through  two  ana  three  points.  ..  .144-161 

CHAPTER  IX. 
Influence  Lines, 

Definition,    influence   lines   for   determining   reactions   shears   arid   moments   on 

simple  beams  and  trusses  and  stresses  in  trusses 162-171 

CHAPTER  X. 
Desig-n  of  I-Beams   and  Plate  Girders. 

Data  and  method  of  procedure,  section  modulus,  economic  depth  of  girders, 
designing  of  flanges  and  webs,  length  of  cover  plates,  flange  increments, 
rivet  spacing  in  flanges,  web  splice,  stiffening  angles,  and  examples 172-104 


CHAPTER  XL 
Design  of  Simple  Railroad  Bridges. 

Types,  loadings,  design  of  beam,  plate  girder  and  truss  bridges  and  viaducts  and 

deflection  and  camber  of  trusses    .  .  195-532 


CHAPTER  XII. 
Design  of  Simple  Highway  Bridges. 

Types,  loadings,  specifications  and  design  of  beam,  pony  truss  and  high  truss 

bridges     583-569 


CHAPTER  XIII. 

Skew  Bridges,  Bridges  on  Curve,  Economic  Height  and  Length  of  Trusses  and 

Stresses  in  Portals. 

Types,  determination  of  stresses  in  skew  bridges,  determination  of  stresses  due 
to  centrifugal  force,  economic  depth  and  length  of  ordinary  trusses  and 
determination  of  stresses  in  various  types  of  portals  570-582 


CHAPTER  XIV. 
Design  of  Buildings. 

Description,  loadings,  and  designing  of  various  types  of  mill  buildings  and  the 

analysis  and   design   of  high   buildings    583-652 


PREFACE 

This  book  is  intended  as  a  textbook  for  college  students  and  as  a 
self-explanatory  manual  of  structural  engineering  for  practical  men. 

During  the  author's  nineteen  years  of  engineering  experience  (the 
greater  part  of  which  was  spent  in  actual  practice)  it  fell  to  his  lot  to 
"break  in"  students  from  most  all  of  our  engineering  schools,  and  he  has 
no  apology  to  offer  for  the  elementary  mechanics  given  in  this  volume,  as 
he  is  convinced  that  no  matter  how  thorough  the  course  in  mathematics 
and  theoretical  mechanics  may  be  it  is  quite  desirable  that  students  have 
a  short  review  of  the  materialistic  phase  of  mechanics  at  the  beginning 
of  the  subject  of  structures  to  whet  their  appetites  for  the  work. 

As  regards  the  designing  given,  the  author  has  endeavored  to  present 
the  usual  methods,  and  this  in  such  a  fashion  that  the  average  engineering 
student  as  well  as  the  practical  man  can  read,  understandingly,  without 
having  a  dictionary,  glossary,  or  a  compendium  on  theoretical  mechanics 
at  his  elbow. 

The  author's  aim  has  been  to  arrange  for  college  use  the  drawing 
room  exercises  throughout  the  text,  so  that  the  work  in  the  classroom 
and  drawing  room  will  go  hand  in  hand.  The  appreciation  of  this 
feature  will  depend  a  great  deal  upon  the  allotment  of  hours  for  the  two 
classes  of  work.  The  ratio  of  two  hours  recitation  to  six  in  drawing  is 
recommended.  "Drawing  Room  Exercise  No.  1,"  at  the  end  of  Chapter 
II,  can  be  started  at  the  very  first  drawing  room  period  without  wasting 
any  time.  Chapters  I,  II,  VII,  X,  and  Chapter  XI  as  far  as  "Deck 
Plate  Girder  Bridges,"  should  be  read  by  the  time  "Drawing  Room 
Exercise  No.  1"  is  finished,  so  that  "Drawing  Room  Exercise  No.  2"  can 
be  taken  up.  During  the  time  spent  on  "Drawing  Room  Exercises  No.  2, 
No.  3,  and  No.  4"  ample  time  will  be  found  for  the  necessary  advance 
reading — up  to  "Through  Plate  Girder  Bridges" — and  the  reading  of 
Chapters  III  to  VI  and  also  VIII  and  IX.  Beyond  this  the  author 
refrains  from  making  suggestions  as  to  assignments  as  he  has  confidence 
in  the  ability  of  instructors  to  assign  correctly  the  work  to  suit  conditions. 

The  practical  men  of  limited  theoretical  training  and  others  desiring 
a  review  of  the  subject  treated,  will  find  the  book  well  suited  to  their  case 
if  the  chapters  be  taken  up  in  consecutive  order. 

The  designs  given  in  this  book  are  entirely  the  work  of  the  author 
and  are  designed  especially  for  this  work,  yet  he  has  endeavored  to  make 
them  as  general  as  possible. 

This  book,  which  treats  only  of  simple  structures,  is  the  author's 
first  volume  on  Structural  Engineering.  A  second  volume,  which  will 
be  known  as  Higher  Structures,  is  in  preparation;  this  will  treat  of 
Movable  Bridges,  Cantilever,  Arch,  and  Suspension  Bridges,  Secondary 
Stresses,  etc. 

The  author  desires  to  acknowledge  his  appreciation  of  the  work  of 
his  assistant,  Mr.  B.  S.  Myers,  in  reading  and  checking  proof  and  pre- 
paring and  checking  drawings,  and  to  thank  Prof.  F.  O.  Dufour  for 
valuable  suggestions  and  criticisms.  J.  E.  KIRKHAM. 

Ames,  Iowa,  Sept.  5,  1914. 


CHAPTER  I 

PRELIMINARY 

1.  Structural  Engineering. — The  part  of  Civil  Engineering  per- 
taining to  the  designing  of  steel  structures,  such  as  bridges,  buildings, 
towers,  etc.,  is  known  as   Structural  Engineering.      The  work  involved 
consists,  principally,  in  determining  the  stresses,  selecting  the  material 
to  be  used,  known  as  sections,  and  the  contriving  and  drawing  of  the 
details.     But,  in  addition  to  this  work,  the  structural  engineer  has,  as  a 
rule,  the  designing  of  a  great  deal  of  incidental  construction,  such  as 
foundations,  concrete  floors,  roofs,  etc. 

In  order  to  design  structures  properly,  one  must  be  perfectly 
familiar  with  the  material  used  in  their  construction  and  have  a  clear 
understanding  of  the  manner  in  which  their  manufacture  and  erection 
are  accomplished  as  well  as  have  a  thorough  knowledge  of  the  mechanical 
principles  involved  throughout.  A  properly  designed  structure  is  one 
wherein  no  mechanical  principles  are  seriously  violated  and  which  at  the 
same  time  is  economic  in  material  and  easily  manufactured  and  erected. 

2.  Structural  Material. — Steel,  concrete,  stone,  and  wood  are  the 
principal  materials  used  by  the  structural  engineer  in  the  construction  of 
modern  structures.     However,  cast  iron,  wrought  iron,  and  a  few  other 
materials  are  used  in  a  few  cases,  as  will  be  designated  when  these  cases 
present  themselves  in  the  text.    Concrete  and  stone  will  not  be  treated  in 
this  book  further  than  to  designate  certain  allowable  pressures. 

3.  Manufacture  Of  Steel. — In   describing   in   a   general   way  the 
usual  process  of  making  steel,  we  can  say  that  steel  is  made  from  iron  ore 
which  is  mined  and  carried  to  a  blast  furnace  through  which  it  is  run 
together  with  coke  and  limestone  and  thus  converted  into  cast  iron.     The 
cast  iron  is  then  taken  to  either  an  open-hearth  furnace  or  to  a  Bessemer 
furnace,  usually  spoken  of  as  a  Bessemer  converter,  and  there  converted 
into  steel.     When  the  conversion  is  complete,  the  molten  steel  is  run  into 
ingot   molds,   which   are   rectangular   cast-iron  molds,   and   molded   into 
ingots.     These  ingots  are  allowed  to  cool  to  some  extent.     Then  the  molds 
are  pulled  off  and  the  ingots  are  taken  to  a  rolling  mill,  which  is  known 
as  a  slab  or  blooming  mill,  where  they  are  heated  in  what  is  known  as  a 
soaking  pit  until  they  have  the  proper  temperature  for  rolling.     Then 
they  are  rolled  to  a  convenient  rectangular  cross-section  and  cut  up  into 
convenient-sized  pieces  known  as  slabs  or  billets.     These  billets  or  slabs, 
as  the  case  may  be,  are  then  taken  to  another  rolling  mill  and  reheated 
very  much  the  same  as  the  ingots  just  described,  and  then  rolled  into 
structural  shapes,  plates,  rails,  etc.,  ready  for  the  market. 

Sometimes  the  cast  iron  from  the  blast  furnace  is  cast  into  rough 
bars,  known  as  pig  iron,  which  may  be  stored  and  converted  into  steel 
or  molded  into  castings  at  any  desired  time,  or  it  may  be  shipped  to  some 
distant  point  to  be  utilized  in  the  same  manner.  At  the  most  modern 
plants  the  molten  metal  is  taken  directly  from  the  blast  furnace  and 
converted  into  steel  without  letting  it  cool  to  any  great  extent. 

1 


2  STRUCTURAL  ENGINEERING 

The  steel  produced  in  an  open-hearth  furnace  is  known  as  "open- 
hearth  steel/'  while  the  steel  produced  in  a  Bessemer  converter  is  known 
as  "Bessemer  steel."  Open-hearth  steel  is  considered  more  reliable  in 
every  respect  than  Bessemer,,  and,  consequently,  the  Bessemer  steel  is 
being  fast  replaced  by  the  open-hearth  product.  In  fact,  most  engineers 
at  present  exclude  the  use  of  Bessemer  steel  in  structural  work  altogether. 

4.  Grades  of  Common  Structural  Steel. —  There  are  three  recog- 
nized grades   of  common   structural   steel:      Medium   Steel;   Soft   Steel; 
Rivet  Steel. 

Medium  Steel  is  a  medium-hard  steel  which  has  practically  replaced 
all  other  grades  of  steel  in  structural  work.  It  has  an  ultimate  strength 
of  60,000  to  70,000  pounds  per  square  inch,  and  an  elastic  limit  of  30,000 
to  35,000  pounds  per  square  inch. 

Soft  Steel  is  softer  than  medium  steel.  It  was  the  grade  of  steel 
first  used  in  structural  work,  but  has  been  practically  replaced  by  medium 
steel.  It  has  an  ultimate  strength  of  52,000  to  62,000  pounds  per  square 
inch  and  an  elastic  limit  of  26,000  to  31,000  pounds  per  square  inch. 

Rivet  Steel  is  a  very  soft  steel  used  almost  exclusively  for  making 
rivets  and  bolts.  It  has  about  the  same  chemical  composition  as  wrought 
iron,  but  has  a  higher  ultimate  strength  and  elastic  limit.  It  has  an 
ultimate  strength  of  48,000  to  58,000  pounds  per  square  inch  and  an 
elastic  limit  of  24,000  to  29,000  pounds  per  square  inch. 

5.  Nickel  Steel  is  a  steel  containing  from  2  per  cent  to  3J  per  cent 
of  nickel.     It  has  an  ultimate  strength  of  80,000  to  112,000  pounds  per 
square  inch  and  has   a  very   high   elastic  limit — something  like   60,000 
pounds  per  square  inch.     This  metal  has  been  used  recently  in  some  of 
the  large  bridges  in  this  country. 

6.  High  Carbon  Steel  is  a  hard  steel  which  contains  more  com- 
bined carbon  than  the  ordinary  medium  steel.     It  has  practically  as  high 
ultimate  strength  and  elastic  limit  as  nickel  steel.     It  is  used  almost  exclu- 
sively in  the  form  of  rods  to  reinforce  concrete. 

7.  Wood,  or  timber,  as  it  is  usually  spoken  of,  is  used  in  structural 
work  principally  for  floor  and  roof  covering.     The  timber  mostly  used  is 
white  oak  and  yellow  pine,  both  of  which  have  an  average  ultimate  crush- 
ing strength   of  about   7,000  pounds   per  square  inch,  and   an  ultimate 
tensile  strength  of  about  twice  that  amount. 

8.  A  Casting  is  made  by  running  molten  metal  into  an  impression 
made  in  sand  by  means  of  a  wooden  pattern  which  is  shaped  to  the  size  of 
the  metal  piece  desired.     The  castings  used  in  structural  work  are  made 
of  either  cast  iron  or  steel.     The  castings  made  of  steel  are  much  stronger 
than  those  made  of  cast  iron  but  cost  more.     The  steel  used  for  making 
castings  has  a  somewhat  different  chemical  composition  than  that  of  rolled 
steel,  referred  to  above.     It  is  produced  in  the  same  way,  however,  the 
chemical  composition  being  controlled  mostly  in  the  mixing  of  the  charge. 

9.  Structural  Steel  Shapes. — The  I-beam,  channel,  angle,  Z-bar, 
and  T-shape,  shown  in  Fig.    1,  are  the  principal  steel  shapes  used  in 
structural  work.     The  T-shape  is  not  used  very  extensively  except  for 
very  special  work.      Plates,  while  they  are  not  classified  as  structural, 
shapes,  are  really  used  in  a  more  general  way  than  the  shapes. 


PRELIMINARY 


Web 


I- Beam 


Channel 


The  tables  in  the  back  of  this  book  give  the  properties,  gauges,  etc., 
of  the  shapes  and  plates  in  general  use.  The  use  of  these  tables  will  be 
explained  as  the  occasion  for  their 
use  occurs.  Either  a  Carnegie  or 
Cambria  handbook  is  a  convenient 
book  to  have — in  fact,  practically 
indispensable  in  structural  work 
— but  the  student  must  bear  in 
mind  that  he  is  not  at  liberty  to 
use  just  any  section  therein  that 
he  happens  to  pick  out,  as  a  great 
many  specials  which  are  not  in 
general  use  are  listed,  which  will 
be  furnished  only  when  the  order 
for  such  becomes  large  enough  to 
warrant  the  rolling,  and,  conse- 
quently, in  the  case  of  ordinary 
orders  containing  these  special 
sections,  an  unusual  delay  in 
obtaining  the  material  will  likely 
be  experienced. 

10.    Rivets  and  Riveting. — 

Rivets  are  used  in  structural  work 

to  connect  the  structural  shapes  and  plates  together.     They  are  simple  bits 

of  metal  in  themselves,  but,  indirectly,  they  are  the  source  of  a  great  deal 

.     Length 


Shank 


Shank 


Button  Headftive/- 


Countersunk  /-/eaa '/?/  vef' 

Fig.  2 


of  trouble.  Practically,  there  are  but  two  kinds:  the  buttonhead  rivet 
which  has  a  hemispherical  head,  and  the  countersunk  rivet  which  lias  a 
countersunk  head.  Each  kind  is  shown  in  Fig.  2,  including  the  names  of 
their  parts.  The  buttonhead  rivet  is  the  kind  most  used;  in  fact,  the 
countersunk  rivet  is  never  used  except  where  it  is  absolutely  necessary 
to  do  so. 

As  a  rule,  each  structural  company  manufactures  its  own  rivets. 
They  are  made  by  feeding  red-hot  rods  into  a  machine  which  cuts  the  rods 
into  pieces  of  proper  length  and  at  the  same  time  forms  a  head  on  each 
piece,  thus  completing  the  rivets  as  they  are  shown  in  Fig.  2.  The  way 
in  which  this  is  done  can  be  seen  by  referring  to  Fig.  3,  where  the  parts 
of  the  machine  that  directly  form  the  rivets  are  shown.  Let  us  first 
consider  the  plan  diagram  at  (a),  where  A  represents  the  "header," 
B  a  fixed  die,  C  a  moving  die,  D  a  forked  bar,  and  S  a  bar  known  as  the 
gauge.  While  the  parts  are  in  the  position  shown  at  (a)  the  heated  rod 
is  pushed  into  the  machine,  passing  through  the  slot  in  the  bar  D  until  it 
comes  in  contact  with  the  gauge  S,  as  indicated  by  the  dotted  outline 
e-n-o-k.  Then  the  die  C  moves  toward  B  and  cuts  the  rod  off  at  o  by 


STRUCTURAL  ENGINEERING 


shearing  with  the  bar  D,  and  at  the  same  time  the  piece  of  the  rod  thus 
cut  off  is  caught  in  the  cylindrical  grooves  in  the  dies  (see  section  m-m) 
and  held  firmly  while  the  gauge  S  moves  up  out  of  the  way  of  the  header 
A  which  then  moves  until  it  ^ 

comes  in  contact  with  the  dies 
B  and  C,  upsetting  the  part  e-n 
of  the  piece  of  the  rod  into  the 


D) 

k 

1 

Is 

n 

c 

m 


0 


A 


(a) 


(b) 


, ,. 


Sechon-mrn. 


Fig.  3 


hemispherical  cup  in  the  header 
A,  thus  forming  the  head  of  the  m  I 
rivet.      The   parts   of  the   ma-  t 
chine  are  then  in  the  position  c 

shown  at  (6).  The  header  A 
and  then  the  die  C  and  the 
gauge  S  move  back  to  the  posi- 
tion shown  at  (a),  while  the 
rivet  just  formed  drops  into  a 
cooling  basin.  Then  to  obtain 
the  next  rivet,  the  rod  is  fed 
into  the  machine  as  before. 
Thus  rivets  are  made  at  the 
rate  of  about  one  per  second. 

The  rivets  formed  by  the  machine  just  described  are  the  buttonhead 
rivets.  However,  countersunk  rivets  can  be  made  on  the  same  machine  by 
replacing  the  header  shown  by  a  header  having  a  plane  end  (without  any 
cup),  and  the  dies  shown  by  dies  which  have  the  cylindrical  groove 
chamfered  out  on  the  end  next  to  the  header  so  as  to  form  the  counter- 
sunk head.  In  that  case  the  heads  are  formed  in  the  dies  instead  of  in 
the  header  as  in  the  case  of  the  buttonhead  rivet. 

After  the  rivet  holes  have  been  either  punched  or  drilled  into  the 
plates  and  shapes  to  be  riveted  together,  and  the  same  have  been 
assembled,  each  in  its  proper  place,  rivets  are  heated,  placed  in  the  holes, 
and  "driven."  The  driving  of  a  rivet  consists  principally  of  forming  a 
head  on  the  plane  end,  usually  the  same  as  the  one  on  the  other  end 
formed  by  the  rivet  machine  just  described. 

Rivets  are  driven  by  machines  known  as  "riveters"  except  in  a  few 
cases  where  it  is  necessary  to  drive  them  "by  hand."  The  riveter,  or 
rivet-driving  machine,  has  two  headers  known  as  tools,  one  being  fixed 
and  the  other  movable,  each  of  which  is  very  similar  to  the  header  used 
in  the  rivet-manufacturing  machine  described  above. 

In  order  to  show  the  working  of 
a  riveter,  let  m  (Fig.  4)  represent  an 
angle  which  is  to  be  riveted  to  a  plate  n. 
The  rivet  holes  are  either  punched  or 
drilled  in  the  two  so  as  to  match.  Then 

Sf- .     n  \\ft    /  rjj      the  plate  and  the  angle  are  placed  to- 

( C//     \H>  '        gether  as  shown,  and  the  driving  of  the 

rivets  proceeds  as  follows:  A  rivet  is 
heated  and  placed  in  one  of  the  holes,  as 
shown  at  (a).  Then  the  fixed  tool  B  of  the  riveter  is  placed  against  the 
head  of  the  rivet  and  the  power  is  applied  (which  may  be  air,  steam  or 
water)  which  moves  the  tool  A  toward  B  whereby  the  projecting  part  of 


Fig.   4 


PRELIMINARY 


m 

T 


(d) 


m 

r 


m 


n 


Fig.    5 


the  rivet  is  upset  into  the  cup  in  the  end  of  A,  thus  forming  the  head  on 
that  side.  When  the  tool  A  moves  until  it  practically  touches  the  plate  n, 
the  rivet  is  fully  driven,  as  shown  at  (6).  The  tool  A  is  then  brought 
back  to  its  former  position,,  and  the  riveter  is  placed  on  the  next  rivet  by 
either  moving  the  pieces  being  riveted  together,  or  by  moving  the  riveter, 
and  the  operation  of  driving  is  repeated,  and  so  on. 

In  case  a  countersunk  rivet  is  to  be  driven,  the  rivet  hole  is  either 
punched  or  drilled  the  same  as  though  a  buttonhead  rivet  were  to  be  used. 
Then  the  hole  is  countersunk,  which  means  that  it  is  reamed  out  cone- 
shaped  on  the  side  where  the  countersunk  head  is  to  come  so  that  the 
head  will  just  fit  in.  The  countersinking  of  the  hole  is  done,  usually, 
with  a  flat,  diamond  point  drill,  as  shown  at  (a),  Fig.  5.  The  driving  of 
a  countersunk  rivet  with  a  riveter  is  very  much  the  same  as  driving  a 
buttonhead  rivet.  For  example, 
suppose  a  rivet  having  a  counter- 
sunk head  is  to  be  used  to  connect 
the  angle  m  to  the  plate  n  as  shown 
at  (6),  Fig.  5,  where  the  counter- 
sunk head  is  on  the  same  side  as 
the  angle.  The  rivet  hole  will  be 
countersunk  in  the  angle  as  shown. 
The  countersunk  rivet  is  heated 
and  placed  in  the  hole  from  the 
same  side  as  the  angle  and 
driven  just  the  same  as  was  explained  above  in  the  case  of  the  button- 
head  rivet,  but  the  tool  held  against  the  countersunk  head  would  be  a 
plane  tool  without  a  cup.  If  the  other  tool  has  a  cup  in  it,  the  head  on 
that  side  would  be  an  ordinary  buttonhead,  as  shown  at  (c),  Fig  5.  If  it 
were  necessary  to  have  a  countersunk  head  on  each  side,  the  hole  would 
be  countersunk  in  the  plate  as  well  as  in  the  angle,  as  shown  at  (d), 
Fig.  5.  The  rivet  before  driving  would  have  a  countersunk  head,  the 
same  as  before,  and  the  driving  would  take  place  in  the  same  manner,  but 
each  of  the  tools  would  have  a  plane  end. 

In  cases  where  it  is  either  impossible  or  impracticable  to  use  the 
riveter,  the  rivets  are  driven  by  hand.  This,  in  the  case  of  a  buttonhead 
rivet,  consists  of  heating  the  rivet  and  placing  it  in  the  hole,  and  while 
the  end  of  a  round  bar  which  contains  a  cup  (known  as  a  dolly  bar)  is 
held  firmly  (by  hand)  against  the  head,  the  projecting  end  of  the  rivet  is 
battered  down  with  a  hammer  and  then  formed  into  a  finished  head  by  a 
heading  hammer  (known  as  a  snap)  which  is  held  against  the  battered 
end  of  the  rivet  and  struck  by  a  sledge.  A  heading  hammer,  or  snap,  is 
really  a  two-faced  hammer  with  a  cup  in  one  end  used  to  form  the  head 
of  the  rivet.  In  the  case  of  a  countersunk  rivet,  the  driving  by  hand  is 
practically  the  same  as  in  the  case  of  the  buttonhead  rivet,  but  the  tools 
used  on  the  countersunk  heads  will  be  plane,  that  is,  without  cups. 

There  are  but  very  few  cases  where  it  is  necessary  to  drive  rivets  by 
hand  in  the  shop,  but  practically  all  of  the  rivets  connecting  the  individual 
members  in  a  structure  to  one  another  are  driven  by  hand  as  the  structure 
is  being  erected.  Such  rivets  are  known  in  structural  engineering  as 
"field  rivets."  Holes  left  open  in  the  shop  for  such  rivets  are  known  as 
open  holes  or  field  holes, 


6  STRUCTURAL  ENGINEERING 

The  thickness  of  metal  connected  by  a  rivet  is  known  as  the  grip  of 
the  rivet.  The  diameter  of  the  shank  of  a  rivet  is  known  as  the  diameter 
of  the  rivet.  It  is  also  known  as  the  size  of  the  rivet. 

The  part  of  the  rivet  projecting  beyond  the  material  to  be  riveted 
together  is  just  a  little  longer  than  that  necessary  to  form  a  head,,  for,  in 
driving,  some  of  this  is  taken  up  in  upsetting  the  shank  of  the  rivet  into 
the  hole  which  is  always  about  -fa  in.  larger  in  diameter  than  the  shank. 

The  size  of  the  rivets  used  depends  upon  the  material  to  be  connected. 
A  common  rule  is  that  no  rivet  shall  be  less  in  diameter  than  the  thickness 
of  any  single  piece  connected.  Rivets  vary  in  size  from  J  in.  in  diameter 
to  1£  ins.  The  rivets  most  often  used  are  the  J-  and  J-in.  diameter. 

11.  Method  of  Procedure  in  the  Designing  of  Steel  Structures. 
— As  a  rule,  parties  desiring  structures  built  specify  the  location  and  pur- 
pose, and  limit  to  some  extent  the  amount  they  shall  cost.    The  engineering 
work  really  begins  with  the  survey  of  the  site,  which,  however,  does  not 
necessarily  require  the  services  of  a  structural  engineer,  yet  a  structural 
engineer  should  know  how  to  do  such  work  as  it  is  sometimes  required  of 
him ;  and  also  such  knowledge  is  often  essential  in  working  out  the  design 
of  a  structure.     The  structural  engineering  work  really  begins  after  the 
drawings  showing  the  site  are  made  from  which  the  structural  engineer 
works  out  the  preliminary  plans  of  the  structure.      These  preliminary 
plans  show,  in  a  general  way,  what  is  desired.     They  usually  consist  of 
what  are  known  as  "Stress  Sheets"  and  "General  Drawings."     A  Stress 
Sheet  is   usually   a   single-line   drawing  of  the   structure   wherein   each 
member  is  represented  by  a  single  line  upon  which  the  corresponding 
stresses  and  sections  are  written,  and  the  general  dimensions  of  the  struc- 
ture are  given  in  reference  to  centers  of  bearing  and  to  centers  of  gravity. 
A  General  Drawing  is  usually  intended  to  show  the  structure  as  a  whole. 
In  many  cases  it  is  a  general  picture  of  the  structure  wherein  the  details 
are  shown  in  a  general  way  but  not  fully  dimensioned.     In  the  case  of 
very  important  structures,  the  general  drawings  usually  include  a  general 
picture  of  the  proposed  structure  showing  no  specific  details  at  all.     Such 
drawings   may  be   prepared   by   an   engineer  or  by   an   architect.      The 
purpose  is  to  present  the  general  appearance  of  the  structure.     That  such 
drawings  are  included  does  not  lessen  in  the  least  the  necessity  of  the 
other  general  drawings. 

After  the  general  drawings  have  been  completed  to  the  satisfaction 
of  all  parties  concerned,  they  are  used  as  a  guide  in  making  the  "Shop 
Drawings"  (sometimes  called  "Working  Drawings")  which  show  the 
spacing  of  rivets  and  all  holes,  cuts,  etc.,  for  each  individual  member  of 
the  structure.  After  these  shop  drawings  are  made  and  thoroughly 
checked  and  are  approved  by  all  parties  concerned,  they  are  sent  to  the 
structural  shops  where  each  member  of  the  structure  is  fabricated  accord- 
ingly. 

All  of  the  drawings  referred  to  above  are,  as  a  rule,  made  on  tracing 
cloth  and  blueprints  are  made  from  these  which  are  furnished  to  all 
parties  concerned  instead  of  actual  drawings.  These  prints,  as  a  rule, 
are  what  are  referred  to  as  drawings. 

12.  A  Bridge  Company  is  an  organization  which  designs,  manu- 
factures and  erects  structural  work.    However,  the  name  is  appropriated 


PRELIMINARY  7 

by  various  concerns  which  in  many  cases  are  capahle  of  doing  only  part 
of  this  work.  A  full  organization  really  consists  of  an  operating  depart- 
ment and  an  engineering  department.  The  operating  department  has  the 
general  management  of  the  company,,  especially  the  commercial  end,, 
while  the  engineering  department  attends  to  everything  pertaining  to  the 
engineering.  It  suffices  here  for  us  to  consider  only  the  engineering 
department.  This  department  consists  of  a  designing  and  estimating 
department,  a  draughting  department,  a  shop  organization,  and  an  erect- 
ing department. 

The  work  of  the  designing  and  estimating  department  consists  mainly 
in  determining  the  stresses  in  structures,  the  selecting  of  the  required 
sections,  drawing  up  the  stress  sheets  showing  the  same,  making  prelimi- 
nary estimates  of  the  weight  of  the  material  to  be  used,  and  computing 
the  cost  of  the  work  proposed  to  be  manufactured  by  the  company. 

The  work  of  the  draughting  department  consists  of  the  working  out 
of  the  complete  details  of  structures,  using  the  stresses  and  sections  as 
specified  on  the  stress  sheets  (which  are  furnished  by  either  the  designing 
and  estimating  department  of  the  company  or  by  outside  parties)  and 
making  the  necessary  general  and  shop  drawings.  In  addition,  the 
draughting  department  makes  out  bills  of  the  material  required  from 
which  the  material  is  ordered  from  the  rolling  mills.  These  bills  when 
completed  are  known  as  "Shop  Bills"  and  are  sent  to  the  shops  along 
with  the  shop  drawings. 

The  work  of  the  shop  organization  consists  in  fabricating  the 
individual  members  of  the  structures  in  the  shops  according  to  the  shop 
drawings  furnished  by  the  draughting  department. 

The  work  of  the  erecting  department  consists  in  erecting  the 
structures  in  their  final  position  after  the  individual  members  are  fully 
fabricated  in  the  shops. 

The  shops  are  divided  into  departments  which  are  referred  to  as 
separate  shops.  These  different  shops,  or  departments,  are  as  follows : 
Templet  Shop,  Pattern  Shop,  Laying-off  Shop,  Punch  Shop,  Rivet  Shop, 
Finishing  Shop,  Forge  Shop,  Machine  Shop,  Paint  Shop,  Foundry,  etc., 
the  name  of  each  clearly  indicating  the  nature  of  the  work  done  therein. 
Some  departments  occupy  separate  buildings,  while  in  some  cases  several 
departments  are  under  one  roof. 

13.  Fabrication  of  Steel  Structures. — The  draughting  depart- 
ment of  a  bridge  company  usually  orders  the  material  for  any  structure 
that  the  company  is  to  fabricate  just  as  soon  as  the  preliminary  work  on 
the  shop  drawings  is  far  enough  along.  The  steel  mills  roll  this  material 
and  ship  it  to  the  shops  where  it  is  unloaded  and  placed  in  the  "receiving 
yards,"  where  it  remains  until  the  shops  need  it.  After  the  shop  drawings 
are  completed,  blueprints  are  made  of  them  which  are  sent  to  the  shops, 
each  department  receiving  the  prints  showing  the  part  of  the  work 
required  of  it.  The  fabrication  of  riveted  work,  as  a  rule,  begins  in  the 
templet  shop  where  full-sized  wooden  templets  are  made  for  most  of  the 
shapes  and  plates  in  the  structure.  These  templets  are  made  so  that  each 
hole  and  cut  shown  on  the  shop  drawings  can  be  located  on  the  actual 
pieces  of  metal  used.  These  templets,  as  a  rule,  are  made  of  one-inch 
white  pine  boards,  well  seasoned.  The  templets  are  either  made  of  one 


g  STRUCTURAL  ENGINEERING 

board  or  of  two  or  more  boards  connected  together,  quite  often  forming 
a  frame.  Pasteboard  is  being  used  of  late,  to  some  extent,  in  making 
templets. 

The  templets  are  taken  to  the  laying-off  shop  where  they  are  clamped 
to  the  material  for  which  they  were  made  to  lay  off.  Then  the  workmen 
mark  where  each  hole  is  to  be  in  the  metal  by  placing  a  center  punch  in 
each  hole  in  the  templet  and  striking  the  punch  with  a  hammer,  and 
indicate  the  cuts  required  by  marking  the  outline  of  the  templet  on  the 
metal.  Then  the  templets  are  undamped  and  thrown  to  one  side  and  the 
material  thus  laid  off  is  taken  to  the  shearing  and  punch  shop  where  all 
cuts  are  made  in  accordance  with  the  marks,  and  the  rivet  holes  are 
punched  or  drilled  as  indicated.  After  this  work  is  completed,  the 
material  is  taken  to  the  assembling  shop  where  the  pieces  are  assembled 
so  as  to  form  individual  members  of  the  structure  and  bolted  together 
temporarily,  just  a  few  bolts  being  used  in  each  member.  Then  these 
members  are  taken  to  the  rivet  shop  where  the  rivet  holes  are  reamed,  if 
such  is  called  for,  and  all  the  rivets  indicated  on  the  shop  drawings  are 
driven.  Then  the  members  are  taken  to  the  finishing  shop  where  pin  holes 
are  bored  and  all  surfaces  and  joints  are  finished  as  called  for  on  the  shop 
drawings.  When  this  work  is  completed  the  members  are  taken  to  the 
paint  shop  where  they  are  cleaned  and  painted.  Then  they  are  taken  to 
the  loading  yards  where  they  are  loaded  on  cars  for  shipment  to  the  site. 
Thus  the  fabrication  is  completed. 

Usually  when  plates  and  shapes  are  duplicated  a  great  many  times, 
templets  are  not  used.  The  duplicated  pieces,  in  that  case,  are  run 
through  a  multiple  punch  which  is  so  constructed  that  the  operator  can 
punch  the  rivet  holes  by  referring  directly  to  the  shop  drawings. 

In  addition  to  the  riveted  work  referred  to  above,  there  are  usually 
castings,  forged  work  and  machine-shop  work  included  in  each  structure 
which  are  gotten  out  by  the  foundry,  forge  shop  and  machine  shop 
independently  of  the  other  shops,  except  that  the  patterns  used  in 
molding  the  castings  are  made  in  the  pattern  shop,  which  is  usually  very 
closely  associated  with  the  templet  shop.  This  work,  when  completed,  is 
loaded  on  cars  and  shipped  to  the  site,  the  same  as  the  riveted  work,  often 
going  on  the  same  cars. 


CHAPTER  II 

STRUCTURAL  DRAUGHTING 

14.  Preliminary. —  It   is   very   essential  that   all  young   engineers 
should   be    good    draughtsmen,,    as   they    are   usually   called    upon   to   do 
considerable    draughting,    and    good    draughting    is    highly    appreciated 
throughout  the  engineering  profession.      Good  draughting  is  an  accom- 
plishment; however,  one  sometimes  hears  of  a  "natural  born"  draughts- 
man.   A  newspaper  reporter  while  talking  to  one  of  our  greatest  inventors 
referred  to  him   as   a   genius.      The   inventor  retorted   that   genius   and 
perspiration    were    synonymous    terms.      The    inventor   may    have    been 
slightly  mistaken   in  that  particular  case,  but  in  this   case  there  is  no 
question:  anyone  can  acquire  the  art  of  making  good  drawings,  that  is, 
good,  neat,  plain,  practical  drawings,  nothing  of  an  artistic  nature  being 
considered,  by  simply  going  about  it  with  a  will,  being  careful,  each  time 
trying  to  do  a  little  better  than  before. 

Structural  drawings  consist  of  lines,  letters,  and  figures.  The  lines 
should  be  true  and  distinct.  The  letters  and  figures  should  be  plain,  neat, 
free-hand  letters  and  figures.  Each  can  be  practiced  independently  of 
the  others,  but  after  a  fair  amount  of  such  practice,  good  draughtsmanship 
can  be  most  readily  acquired  by  making  exact  copies  of  some  good  draw- 
ings, providing  the  letters  and  figures  be  of  the  same  type  as  those  being 
practiced  by  the  student,  as  it  is  not  advisable  to  shift  from  one  type  to 
another  until  one  type  is  fairly  mastered. 

15.  The    Necessary    Equipment    for    Structural    Draughting 

consists  of  a  drawing  board,  T-square,  two  triangles,  one  decimal  scale., 
one  duodecimal  scale,  one  first  class,  medium  size,  right  line  drawing  pen, 
one  first  class,  small  ink  compass,  one  large  combined  ink  and  pencil  com- 
pass, one  pair  of  medium  size  dividers,  a  slide  rule,  either  a  Carnegie  or 
Cambria  handbook,  one  small  bottle  of  black  waterproof  drawing  ink,  a 
writing  pen  and  holder,  a  scratch  pad,  drawing  pencils,  thumb  tacks, 
erasers,  drawing  paper,  and  tracing  cloth. 

There  are  several  very  useful  things  which  could  be  added  to  the 
above  list,  such  as  logarithmic  tables,  book  of  squares,  beam  compass,  etc. 
There  are  no  objections  to  a  full  set  of  drawing  instruments  instead  of  the 
few  instruments  mentioned  above,  but  in  either  case,  the  instruments 
should  be  first  class.  It  is  better  to  have  a  few  good  instruments  than  a 
full  set  of  poor  ones. 

16.  Free-Hand  Letters  and  Figures. — Some  distinct  character- 
istics will  always  be  seen  in  the  free-hand  lettering  of  each  individual  the 
same  as  in  the  case  of  ordinary  writing,  but  the  type  of  letters  and  figures 
should   always   conform   to   some   recognized   standard.      Poor   lettering 
should  not  be  recognized  as  a  characteristic,  but  as  an  indication  of  lack 
of  practice. 

9 


10  STRUCTURAL  ENGINEERING 

Most  of  the  free-hand  lettering  on  drawings  is  done  on  tracing  cloth, 
so  it  is  best  to  practice  upon  the  same.  The  small  remnants  of  cloth 
which  would  otherwise  be  wasted  can  often  be  utilized  for  that  purpose. 
The  student  can  have  the  necessary  material,  which  includes  a  small 

£y$fem  of  Lettering 

SmaJI  LeM&ns 
a(c>) 


c 


s(s) 

Figures 


/2  t  2         '  23  '* 


/  un/ess 

nof-ecf.  /?//  rivets  soft  sfee  /  ar>c/  -j  '  c//a. 
All  rivet  ho/es  punch&c/  ' 


for 
2-2SO-0  "5wg/e  Track  7ft  ra  fto  Con.  Spans 


Jfo. 

Figr-   6 

bottle  of  drawing  ink,  writing  pen,  and  tracing  cloth,  at  his  private  room 
and  practice  during  spare  moments  as  a  diversion.  Opinions  will  diffei 
somewhat  as  to  the  kind  of  pen  to  use  in  making  free-hand  letters  and 
figures.  The  author  prefers  Gillott's  No.  303. 

The  letters  and  figures  shown  in  Fig.  6  are  of  about  the  type  used  on 
structural  drawings.     Just  how  each  letter  and  figure  is  made  is   fullv 
indicated,  the  arrows  indicating  the  direction  of 
the    strokes,    and    the    small    figures    above    the 
letters  and  figures  indicating  the  order  in  which 

the  strokes  are  made. 

Fig.  7  The   student  can   best   practice   making   the 

letters  and  figures  shown  in  Fig.  6  by  making  as 

nearly  as  possible  exact  copies  of  the  work  shown  there.  In  addition,  at 
intervals  he  should  practice  drawing  parallel  lines  (free  hand)  having  the 
same  slope  as  the  letters,  and  curves  which  form  the  letter  O,  all  of  which 
is  outlined  in  Fig.  7.  If  there  is  any  letter  or  figure  which  gives  the 


STRUCTURAL  DRAUGHTING 


11 


student  particular  trouble,  he  should  keep  practicing  upon  it  at  intervals 
until  he  has  mastered  it. 

17.  Size  of  Drawings.— The  usual  practice  in  structural  work  is 
to  make  the  drawings  23  x  35  inches  inside  the  border  line  with  a  one-half 
inch  margin  on  all  sides.     This,  however,  is  not  an  absolutely  fixed  size. 
In  some  cases  it  is  necessary  to  make  larger  drawings,  and  in  other  cases 
it  is  convenient  to  make  smaller  ones.     Drawings  18  x  24  inches  is  a  very 
convenient  size  in  some  cases.     The  border  line  should  be  an  ordinary 
plain,  single  line. 

18.  Tracings. —  It  is  the  usual  practice  to  pencil  out  all  drawings 
upon  ordinary  drawing  paper,  and  then  make  tracings  of  these  pencil 
drawings  upon  tracing  cloth.     The  drawing  on  the  tracing  cloth  should  be 
upon  the   dull  side.      Before  starting  the  tracing,   the  cloth  should  be 
rubbed  over  with  talcum  powder  or  with  chalk,  which  should  then  be 
brushed  off  with  a  cloth  or  brush.     A  knife  should  never  be  used  to  make 
erasures    on   the    cloth.      Repeated   erasures    can   be   made    by    using   a 
comparatively  soft  rubber,  as  the  Ruby  Eberhard  Faber  No.  112.     No 
other  than  this  quality  of  eraser  should  be  used. 

19.  General  Hints  Regarding  Shop  Drawings.— The  shop  draw- 
ings for  a  structure  are  usually  the  last  drawings  made.,  but  in  order  to 
work  up  the  preliminary  drawings  satisfactorily,  .knowledge  of  shop  draw- 
ings is  indispensable,  which  makes  it  necessary  that  the  student  take  up  the 
making  of  some  simple  shop  drawings  as  preliminary  work.     As  stated 
above,  the  shop  drawings  show  the  complete  details   of  the  individual 
members  of  structures,  that  is,  all  rivets,  holes,  and  cuts  are  located,  and 
sizes  of  all  shapes  and  plates  are  given,  and  in  addition,  the  kinds  and 


SHOP  R/YETS 


HELD  R/VET3 


" 


P/o/n  Countersunk  F/a//ened  fog 


<fc 


Countersunk  Plain 


ll.  ,«  '^    «:•§-:« 


Fig.   8 

sizes  of  rivets  and  holes  and  also  the  nature  of  the  machine  work  desired 
are  indicated. 

In  order  to  indicate  the  kind  of  rivets  desired,  certain  conventional 
signs  are  used,  most  of  which  are  shown  in  Fig.  8.  Shop  rivets  are  those 
driven  by  riveters  while  the  members  are  in  the  shops,  and  the  field  rivets 
are  those  driven  at  the  site  as  the  structures  are  being  erected.  Counter- 
sunk rivets  are  used  on  the  account  of  clearance,  but  sometimes  sufficient 


12 


STRUCTURAL  ENGINEERING 


clearance  will  be  obtained  by  just  mashing  down  the  heads  of  the  rivets. 
In  such  cases  the  heads  are  said  to  be  flattened. 

Rivet  holes  are  usually  -f%  inch  larger  than  the  rivets  to  be  driven 
into  them.  The  sizes  of  the  heads  of  rivets  are  given  in  Fig.  9.  The 
sizes  of  the  heads  shown  on  drawings  should  correspond  with  the  sizes 
given  here. 

In  locating  rivets,  the  following  rules  should  be  practically  followed : 
Never  space  rivets  closer  together,  that  is,  center  to  center,  than  three 


!L" 

I6_ 

icJB?* 


Ib 


m 


i-iier/ \  'IMF/    \ 

\    ,2.\         I    ,9,,\ 


\./8\ 


Fig.   9 


times  the  diameter  of  the  rivets,  nor  farther  apart  than  six  inches.  Never 
space  rivets  closer  to  the  edge  or  end  of  a  shape  or  plate  than  two  times 
their  diameter.  These  rules  are  not  iron-clad,  but  the  deviation  from 
them  should  be  slight.  For  example :  in  practice  it  is  customary  to  space 
J-inch  rivets  1^  inches  from  the  edge  or  end  of  a  shape  or  plate,  and  f-inch 
rivetis  1^  inches.  It  is  also  customary  to  limit  the  minimum  distance 
between  f-inch  rivets  to  3  inches,  and  f-inch  rivets  to  2J  inches. 

The  spacing  of  rivets  or  holes  in  reference  to  one  another  along  one 
direction  is  known  as  the  pitch.  Rivets  or  holes  passing  through  the 
flange  of  any  shape  are  located  in  reference  to  the  shape  by  what  is  known 
as  the  gauge.  For  example,  the  distance  between  any  two  consecutive 
rivets  or  holes  along  the  angles  shown  in  Fig.  10  is  the  pitch  at  that 
point,  while  the  distance  g  from  the  back  of  the  angles  to  the  line  o-o, 
passing  through  the  rivets,  is  the  gauge.  The  distances  marked  e  are 


Fig.   10 

known  as  the  end  distances.  In  case  the  flange  of  a  shape  has  double 
gauge  lines,  the  pitch  is  the  distance  between  the  rivets  measured  along 
the  piece  and  not  the  distance  between  the  rivets  on  one  gauge  line. 

The  standard  gauges  for  shapes  are  given  in  the  tables  in  the  back 
of  this  book.  The  same  are  to  be  found  in  the  various  "standards"  and 
handbooks  gotten  out  by  structural  companies. 

No  rivet  or  hole  should  be  located  in  reference  to  the  edge  of  a 
flange,  as  would  be  done  by  giving  the  distance  x  in  Fig.  10. 

All  lines  upon  which  rivets  are  located  are  known,  in  general,  as 
rivet  lines,  and  the  lines  used  in  giving  the  spacing  of  the  rivets,  as  well 
as  all  other  distances,  are  known  as  dimension  lines.  All  rivet  lines 
(including  gauge  lines  and  dimension  lines)  should  be  light  in  comparison 


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STRUCTURAL  DRAUGHTING  13 

with  the  lines  used  to  outline  the  members.     All  should  be  in  black  ink. 

The  duo-decimal  scale  is  used  in  laying  out  structural  drawings. 
The  usual  scale  for  shop  drawings  is  three-quarters  inch  equals  one  foot, 
but  in  some  cases  a  scale  of  one-half  inch  equals  one  foot  is  used.  In 
light  work  it  is  sometimes  advisable  to  use  a  larger  scale,  in  which  case 
a  scale  of  one  inch  equals  one  foot  or  one  and  one-half  inches  equals  one 
foot  is  used.  It  is  advisable  for  the  student  to  use  a  large  scale  at  the 
beginning. 

In  making  structural  drawings,  it  is  customary  to  represent  feet  by 
putting  one  dot  above  the  figures  specifying  the  number  of  feet,  and  two 
dots  in  the  case  of  inches.  Thus  4'  is  read  as  four  feet,  and  4"  is  read  as 
four  inches.  2' -2"  is  read  as  two  feet  and  two  inches;  6'-7-|"  is  read 
as  six  feet  and  seven  and  one-half  inches,  and  so  on.  Square  feet  and 
square  inches  are  represented  by  placing  a  square  in  front  of  the  dots. 
Thus  4a'  is  read  as  four  square  feet,  and  4^"  is  read  as  four  square 
inches. 

In  specifying  the  material  on  structural  drawings  it  is  customary  to 
represent  the  shapes  by  symbols  instead  of  writing  the  name.  Thus,  the 
I-beam,  channel,  Z-bar,  and  angle  are  represented  by  I,  [,  Z,  and  L, 
respectively,  or  in  case  of  plural,  as  Is,  [s,  Zs,  and  Ls,  respectively. 

All  dimensions  and  sizes  given  on  structural  drawings  are  in  feet 
and  inches  and  fractional  parts  of  inches.  The  fractional  parts  are 
expressed  in  halves,  fourths,  eighths,  sixteenths,  thirty-seconds,  and 
sometimes  in  sixty-fourths  of  an  inch. 

Depths,  widths,  and  thicknesses  are  always  expressed  in  inches, 
while  length  is  always  expressed  in  feet  and  inches. 

In  the  case  of  I-beams  and  channels,  the  number  required,  depth, 
weight  and  length  are  given  thus: 

1— I  12"x30#x20'-4";    1— [  15"x33#x6'-9Ty; 

2— Is  15"x42£x21'-8i";    6— [s  12"  x  35* x  12'-6J";  etc.;  etc. 

In  the  case  of  angles,  the  number  required,  width  of  flanges  (known 
as  legs),  thickness  of  metal,  and  length  are  given,  thus: 

1__L  5"  x  31"  x  i"  x  6'- W ;   1— L  6"  x  6"  x  f  "  x  9'-7l" ; 

4— Ls  V  x  4"  x  J"  x  12'-8i" ;   7— Ls  3"  x  V  x  f  "  x  4'-0" ;  etc. 

In  case  of  plates,  the  number  required,  width,  thickness,  and  length 
are  given,  thus: 

1— Plate  36"XlL"xl4'-2i";   6— Plates   14"  x  Ty  x  4'-l". 

In  detailing  a  member,  it  is  the  usual  practice  to  show  an  elevation, 
top  view,  and  a  section  through  the  member,  looking  downward.  How- 
ever, there  should  always  be  as  many  sections  and  views  taken  as  is 
necessary  to  clearly  show  the  member  in  detail. 

DRAWING  ROOM  EXERCISE  NO.  1 

Reproduce  Plates  1,  2,  and  3  each  to  a  scale  of  1J"  =  1'-0".  Each 
drawing  should  first  be  reproduced  in  pencil  upon  an  18  x  24-inch  sheet 
of  ordinary  drawing  paper,  then  traced  upon  a  sheet  of  tracing  cloth  of 
the  same  size.  In  making  these  drawings,  the  student  should  study  each 
detail  as  he  goes  along.  All  necessary  information  concerning  the 
material  shown  on  these  drawings  will  be  found  in  the  tables  in  the  back 
of  this  book  or  in  a  Carnegie  or  Cambria  handbook. 


CHAPTER  III 

FUNDAMENTAL  ELEMENTS  OF  STRUCTURAL  MECHANICS 

20.  Structural  Mechanics. —  Mechanics,  in  general,  is  the  science 
which  treats  of  the  effects  produced  upon  bodies  by  force,  while  Structural 
Mechanics,  as  the  name  would  imply,  is  simply  that  part  of  mechanics 
that  applies  directly  to  structures. 

21.  Body. — In  common  usage,  the  word  "body"  refers  to  what  is 
known  as  an  ordinary  solid  body.     But  in  reality  we  are  unable  to  think 
of  material   (matter)   at  all  other  than  as  bodies.     This  is  most  readily 
realized  by  attempting  to  conceive  of  material  in  a  rarefied  state,  as  in  the 
composition  of  a  gas.     We  can  conceive  of  it  only  as  being  made  up  of 
very  small  bodies,  known  as  particles,  molecules,  etc.     Neither  can  we 
conceive  of  the  composition  of  solid  bodies  or  liquids  otherwise  if  thought 
of  in  detail.     Such  being  the  case,  it  is  evident  that  any  part  of  a  recog- 
nized body,  to  us,  is  really  a  body  in  itself;  and  as  we  are  not  limited  as  to 
subdivision,  it  is  evident  that  any  part  of  any  body  may,  at  will,  be 
treated  as  an  independent  body. 

22.  Force. — In  mechanics,  the  word  "force"  refers  to  that  which 
we  recognize  as  the  push  or  pull  that  bodies  exert  upon  each  other,  which 
invariably  produces  motion  or  a  tendency  of  motion  of  the  bodies  con- 
cerned. 

Forces  exerted  by  bodies  at  rest  produce  only  a  tendency  to  motion, 
and  are  known  as  static  forces,  and  the  part  of  mechanics  treating  of  such 
forces  is  known  as  Statics.  Forces  exerted  by  moving  bodies  are  known 
as  dynamic  forces,  and  the  part  of  mechanics  treating  of  such  forces  is 
known  as  Dynamics.  Most  of  the  mechanics  known  as  Structural 
Mechanics  can  be  classed  under  the  head  of  Statics. 

Force  is  known  by  various  names,  as  pressure,  load,  stress,  reaction, 
etc.  These  names  are  used  to  indicate  certain  existing  conditions;  how- 
ever, force  is  the  same  wherever  we  find  it. 

23.  Measure  of  Forces. —  Measuring  forces  is  the  same  as  measur- 
ing any  other  thing;  it  is  simply  a  matter  of  comparison,  that  is,  we  com- 
pare forces  with  some  one  force  which  is  taken  as  the  basis.     We  are  all 
familiar  with  the  practice  of  measuring  forces  in  terms  of  the  common 
gravity  unit,  in  which  case  the  pull  of  gravity  upon  a  certain  body  selected 
as  a  standard  is  taken  as  the  basis  of  comparison  and  designated  as  a 
pound  and  known  as  the  unit  of  weight.     It  is  readily  seen  that  this  unit 
is  an  arbitrary  one,  as  the  pull  of  gravity  upon  most  any  body  could  have 
been  selected  as  the  unit,  although  in  most  cases  convenience  would  have 
been  sacrificed. 

In  speaking  of  a  force,  we  say  a  force  of  so  many  pounds,  or  the 
intensity  of  a  force  is  so  many  pounds.  But  what  we  really  mean  is  that 
the  push  or  pull  (as  the  case  may  be)  exerted  upon  a  certain  body  by 

14 


FUNDAMENTAL  ELEMENTS  15 

some  other  body  is  so  many  times  greater  or  less  than  the  pull  of  gravity 
upon  the  standard  body,  which  pull,  as  stated  above,  is  known  as  a  pound. 
It  is  evident  that  any  static  force  is  directly  measurable  as  compara- 
tive weight,  but  dynamic  forces  are  more  readily  determined  by  consider- 
ing the  change  of  motion  produced  by  them.  It  is  known  by  experiment 
that  the  force  of  gravity  will  cause  the  velocity  of  a  body  falling  freely 
to  increase  about  32.2  feet  per  second  for  each  second  of  time.  This 
increment  of  velocity,  32.2  feet,  is  known  as  the  acceleration  of  gravity 
and  is  usually  represented  by  the  letter  "g."  Then  using  the  force  of 
gravity  expressed  in  the  pound  unit  as  our  basis  of  comparison  (as  in 
static  forces)  we  can  say  that  the  intensity  of  any  dynamic  force  is  to  the 
force  of  gravity  as  the  acceleration  produced  by  the  dynamic  force  is  to 
the  acceleration  produced  by  gravity,  regardless  of  whether  the  body 
considered  moves  horizontally,  vertically,  or  in  any  other  direction.  For 
example,  suppose  a  dynamic  force  of  an  unknown  intensity  F  by  acting 
continuously  upon  a  body  weighing  W  pounds  increases  its  velocity  a  feet 
per  second  for  each  second  of  time.  Then  by  direct  proportion  we  have 

F  :  W  ::  a  :  g. 

Expressing  this  in  words  we  say:  The  unknown  force  (F)  in  pounds  is 
to  the  force  of  gravity  (W)  in  pounds  as  the  acceleration  (a)  in  feet  per 
second  caused  by  the  unknown  force  (F)  is  to  the  acceleration  (g)  caused 
by  gravity.  From  which  we  have 

F=j« (A), 

which  is  one  of  the  fundamental  equations  of  mechanics.  The  equivalent 
formula  is  often  written :  F  =  ma,  where  m  =  W  /g,  and  is  known  as  the 
mass  of  the  body  considered.  But  Formula  (A)  is  the  one  mostly  used 
in  practical  work.  The  mass  of  a  body  is  the  quantity  of  material  it 
contains ;  but  in  a  practical  sense,  the  mass  of  a  body  is  equivalent  to  what 
its  weight  would  be  if  the  acceleration  of  gravity  were  one  foot  (consider- 
ing a  foot  as  unity)  per  second  instead  of  32.2  feet, — which  it  really  is. 
Then  by  direct  proportion,,  using  the  force  of  gravity  as  our  basis  as 
before,  we  have 

W 

m  :  W  : :  1  :  g,  or  m  -  — 

t/ 

which  is  the  expression  for  mass  given  above. 

The  above  discussion  refers  to  constant  forces,  as  the  forces  encoun- 
tered in  structural  engineering  are  mostly  constant  forces.  In  case  of 
variable  forces  it  is  necessary  to  know  the  rate  and  the  limit  of  variation 
first,  then  the  intensity  can  be  determined. 

Problem  1.  A  body  weighing  200  Ibs.  is  acted  upon  by  a  constant 
force  which  increases  its  velocity  from  0  to  64.4  ft.  in  1  second  of  time. 
What  is  the  intensity  of  the  force? 

Solution:     Substituting  in  Formula  (A),  we  have 

F  =  ^x  64.4  -400  Ibs. 
32.2 

Problem  2.  What  would  be  the  intensity  of  the  force  in  Problem  1 
if  the  velocity  of  the  body  were  increased  in  5  seconds  from  50  ft.  per 
second  to  150  ft.  per  second? 


16  STRUCTURAL  ENGINEERING 

Solution:      Here  the  increment  of  velocity  or  acceleration  per  second 
is  (150  —  50)  -h  5  =  20  ft.     Then  substituting  in  Formula  (A),  we  have 

20  =  124  Ibs.  (about). 


Problem  3.  If  the  body  in  Problem  1  had  a  velocity  of  150  ft.  per 
second,,  what  would  be  the  intensity  of  a  constant  force  required  to  stop 
the  body  in  3  seconds? 

24.  Direction  of  Action  of  a  Force.  —  Whenever  a  force  is  applied 
to  a  body  it  either  moves  the  body  or  has  a  tendency  to  move  it.     The 
direction  in  which  the  body  moves  or  has  a  tendency  to  move  is  said  to  be 
the  direction  of  action  of  the  force  causing  the  motion  or  tendency. 

25.  Line  of  Action  of  a  Force.  —  Whenever  a  body  receives  a  force 
by  coming  into  contact  with  another  body,  which  is  the  most  common  case, 
there  is  always  a  surface  upon  which  the  force  is  exerted.     WTe  cannot 
conceive  of  this  force  being  transmitted  to  the  surface  other  than  that 
each  infinitesimal  area  of  the  surface  receives  a  small  amount  of  the  total 
force  transmitted.     So  it  is  evident  that  what  we  usually  consider  a  force 
is   really  made  up   of  an  infinite  number  of   forces.      But  it   would  be 
impossible  to  deal  with  forces  in  mechanics  when  considered  in  this  way, 
since  there  would  be  an  infinite  number  of  forces  to  consider  in  every  case. 
We  avoid  this  difficulty  by  assuming  the  total  force  to  be  applied  along  a 
line  which  passes  through  the  surface  at  the  center  of  mean  intensity  of 
the  force  and  in  the  direction  of  its  action.     This  line  is  known  as  the  line 
of  action  of  the  force.    The  line  of  action  of  a  force  is  unlimited  in  length, 
and  the  force  may  be  considered  to  be  applied  at  any  point  along  the  line 
of  action  as  far  as  motion  or  tendency  of  motion  of  the  body  upon  which 
it  acts  is  concerned. 

When  a  force  is  distributed  over  quite  a  large  surface  it  is  usually 
necessary  to  consider  the  surface  divided  up  into  small  portions,  say, 
portions  one  foot  square,  or  one  inch  square,  as  the  case  may  require,  and 
to  treat  the  part  of  the  total  force  exerted  upon  each  portion  as  a  separate 
force.  This  means  that  the  force  exerted  upon  any  portion  will  have  a 
line  of  action  of  its  own  passing  through  that  portion  at  the  center  of  the 
mean  intensity  of  the  force  exerted  upon  it.  In  any  case  we  can  obtain 
only  an  approximation,  but  the  approximation  should  come  within  the 
limits  of  practicability. 

In  case  of  such  forces  as  gravity  and  magnetism,  where  each  particle 
of  material  is  assumed  to  be  equally  affected,  the  line  of  action  passes 
through  what  is  known  as  the  center  of  gravity  of  the  body.  However,  if 
the  bodies  be  large,  as  is  the  usual  case  of  structures,  it  often  becomes 
necessary  to  consider  the  body  to  be  divided  up  into  smaller  ones  and  to 
consider  that  the  force  acting  upon  each  of  the  smaller  bodies  has  a  line 
of  action  of  its  own,  similar  to  the  case  of  surfaces. 

The  plane  in  which  the  line  of  action  of  a  force  lies  is  known  as  the 
plane  of  the  force,  and  the  direction  of  the  line  of  action  is  known  as  the 
direction  of  the  force,  and  the  direction  of  the  action  along  the  line  of 
action  is  known  as  the  direction  of  the  action  of  the  force.  Any  number 
of  forces  could  act  upon  a  body  and  each  force  could  be  in  a  different 
plane,  but  in  most  of  the  problems  in  structural  engineering  the  conditions 
are  such  that  we  can  consider  them  as  acting  in  the  same  plane. 


FUNDAMENTAL  ELEMENTS  17 

26.  Graphical  Indication  of  Forces.— When  considering  the  effects 
produced  upon  a  body  by  forces,  for  convenience  we  graphically  represent 
the  forces  applied  without  reference  to  the  bodies  applying  them.     This  is 
done  by  drawing  ail  arrow  in  the  line  of  action  of  each  force.     The  arrow 
point  in  each  case  indicates  the  direction 

of   action.,   and   the   point   upon   the   body  \<p\  &/  ~j 

where    the    arrow    touches    indicates    the  \  V  ?/ 

point  of  application.     Thus  in  Fig.  11,  the      ^1       5     ~~        b~         c~j  „ 

arrows  PI,  P2,  P3  and  P4  indicate  forces          ~i ' 

applied  upon  the  body  AB  at  the  points  ^/ 

a,  b,  c,  and  d,  respectively.     The  arrows  Fjg   n 

would  be  spoken  of  as  forces  PI,  P2,  etc. 

And  we  would  say  that  the  forces  PI,  P2,  and  P4  act  toward  the  body, 

while  the  force  P3  acts  away  from  it,  this  being  indicated  in  each  case 

by  the  arrow  points. 

27.  Applied  Forces  and  Reactions. — An  applied  force  is  any  force 
which  we  conceive  of  as  being  applied  to  a  body  from  without  by  some 
other  body,  and  at  the  same  time  as  being  the  initiative  of  the  outward 
activity.     A  reaction  is  the  same  as  an  applied  force,  except  we  conceive 
of  it  as  resisting  the  activity  instead  of  being  the  cause  of  it.     Both  are 
spoken  of  as  external  forces.     For  example,  a  body  placed  at  C  upon  the 
bar  AB  (Fig.  12)  by  virtue  of  its  own  weight  will  exert  a  force  P  (pres- 

•  sure)  upon  the  bar  at  that  point.  This 
r force  P  will  cause  the  bar  to  have  a  ten- 

A  Li  *  ijl  B       dency  to  move  downward,  which  is  the  ac- 

JRI  JR2      tivity  produced.  Actual  motion  is  prevented 

Fig.  12  by  the  supports   at  A   and  B  exerting  the 

forces   Rl   and  R2   acting  upward   on  the 

bar.     We  would  call  P  an  applied  force,  as  we  conceive  of  it  as  being 

the  initiative  of  the  activity  of  the  bar,  while  we  would  call  Rl  and  R2 

reactions,  as  we  conceive  of  them  as  resisting  the  activity,  which,  in  this 

case,  as  stated  above,  is  the  tendency  of  motion  caused  by  the  force  P. 

But  regardless  of  name,  all  three  of  these  forces  are  exactly  the  same 

in  character.     This  can  be  seen  very  readily 

by  imagining  the  bar  to  be  turned  upside 

down   and   supported   only  at  C,  while  all 

three  of  the  forces  continue  to  act  the  same 

in    reference   to   the   bar.      Then   the   case 

would  be  as  shown  in  Fig.  13.     This  could  Fig.  13 

be  actually  accomplished  by  placing  a  body 

weighing  the  same  in  pounds  as  the  reaction  Rl  where  Rl  is  indicated 

to  act  and  another  body  weighing  the  same  in  pounds  as  the  reaction  R2 

where  R2  is  indicated  to  act.     Then  Rl  and  R2  would  become  applied 

forces,  while  P  would  become  a  reaction. 

28.  Stress. — Stress  is  the  push  or  pull  which  the  constituent  parts 
of  a  body  exert  upon  each  other  due  to  the  application  of  external  forces. 
Stresses  are  known  as  internal  forces,  while  applied  forces  and  reactions 
are  known  as  external  forces. 

As  a  simple  case,  suppose  a  steel  rod  having  a  uniform  cross-section 
of  A  square  inches  is  suspended  from  one  end  and  a  weight  of  P  pounds 
is  hung  on  the  other.  It  is  known  from  experience  that  the  pull  P  exerted 


18  STRUCTURAL  ENGINEERING 

at  the  lower  end  of  the  rod  by  the  weight  is  in  turn  exerted  at  the  uppe"r 
end.  So  this  pull  is,  as  we  say,  transmitted  through  the  rod,  and  we 
cannot  conceive  of  this  transmission  taking  place  other  than  that  the 
material  particles  exert  a  pull  upon  each  other  in  the  direction  of  the 
length  of  the  rod.  A  practical  idea  of  the  stress  in  the  rod  can  be  obtained 
by  first  imagining  the  weight  P  to  be  supported  by  a  very  small  wire,  so 
small  in  cross-section  that  the  wire  would  be  really  a  row  of  individual 
molecules ;  then  we  can  conceive  of  the  pull  P  being  transmitted  along  the 
wire  from  molecule  to  molecule,  thus  producing  a  stress  of  P  pounds  upon 
each  molecule.  Next  imagine  another  wire  of  the  same  size  suspended 
along  the  side  of  the  wire  just  considered,  and  suppose  this  second  wire 
now  supports  half  of  the  weight  P:  then  the  stress  on  the  molecules  in 
either  wire  would  be  P/2 ;  and  by  adding  another  wire  of  the  same  size 
the  stress  on  the  molecules  in  each  wire  would  be  P/3;  and  by  adding 
another  wire  it  would  be  P/4;  and  so  on.  So  if  there  be  n  number  of 
wires,  the  stress  on  each  would  be  P/n.  Now  imagine  the  rod  made  up 
of  n  number  of  such  wires  and  let  da  be  the  area  of  the  cross-section  of 
each  wire  in  square  inches,  and  let  s  be  the  stress  per  square  inch  on  each 
wire;  also  let  p  be  the  stress  per  square  inch  on  the  rod  at  any  cross- 
section.  It  is  obvious  that  the  stress  per  square  inch  on  the  rod  at  any 
cross-section  will  be  p  =  P/A.  Then,  likewise,  the  stress  on  each  wire 
being  P/n,  the  stress  per  square  inch  in  each  wire  will  be 


s=  — , —  = 


da        nda 

But  nda  =  A; 

therefore,  we  have 

P 

s  =  -=p. 

That  is,  the  stress  on  the  rod  per  square  inch  is  the  same  whether  we 
consider  the  entire  cross-section  or  a  portion  or  a  mere  molecule  in  the 
cross-section.  When  we  say  that  the  stress  on  a  body  is  so  much  per 
square  inch,  we  do  not  necessarily  mean  that  there  is  a  square  inch  of 
material  in  the  body  that  actually  has  that  stress.  We  may  simply  mean 
that  there  is  some  material  in  the  body  that  has  that  stress ;  it  may  be  a 
square  foot  or  a  millionth  part  of  a  square  inch.  The  stress  per  square 
inch  on  a  body  is  known  in  structural  mechanics  as  Unit  Stress. 

If  the  above  rod  were  stood  up  vertically — say,  upon  a  floor — and 
the  weight  placed  upon  the  top  end,  the  direct  stress  produced  by  the 
weight  would  be  the  same  in  intensity  as  if  suspended  as  above,  but  the 
stress  would  be  produced  by  a  push  instead  of  a  pull.  When  the  stress  in 
a  body  is  produced  by  a  direct  pull,  it  is  known  as  Tensile  Stress;  but 
when  it  is  produced  by  a  direct  push,  it  is  known  as  a  Compressive  Stress. 

Whenever  a  body  is  subjected  to  one  kind  of  stress  uniformly  dis- 
tributed over  its  cross-section  throughout  its  length,  as  was  considered  in 
the  case  of  the  above  rod,  the  stress  is  known  as  Simple  Stress.  The  unit 
stress  in  that  case  at  any  cross-section  is  always  equal  to  the  total  stress 
on  the  cross-section  divided  by  the  area  of  the  cross-section,  or,  in  general, 
we  have  p  =  P/A,  as  given  above. 


FUNDAMENTAL  ELEMENTS  19 

From  the  above  discussion  of  stress  one  may  be  led  to  think  of  all 
bodies  as  being  fibrous  in  structure.  It  is  true  thst  wood,  wrought  iron, 
and  some  other  materials  manifest  this  structure  to  some  extent,  but  not 
wholly,  while  some  other  materials,  as  steel  and  concrete  for  example, 
seem  to  be  devoid  of  it.  But  in  any  case  we  .cannot  conceive  of  the 
arrangement  of  the  ultimate  parts  of  a  body  other  than  in  some  consecu- 
tive order,  and  consequently  our  demonstration  is  not  faulty,  as  we  only 
conform  to  that  conception. 

Stress,  known  as  shear,  cross-bending,  and  torsion,  will  be  treated 
later  under  these  specific  heads  as  the  occasion  for  their  treatment  occurs, 

29.  Elasticity. —  It  is  an  observed  fact  that  whenever  a  force  is 
applied  to  a  body,  the  dimensions  of  the  body  are  changed,  and  that  when 
the  force  causing  such  a  change  is  released,  the  body  has  a  tendency  to 
regain  its  original  dimensions,  and  will  regain  them  unless  the  force  be  so 
great  as  to  break  or  injure  the  body.     This  property  of  regaining  their 
original  dimensions  which  bodies  manifest  is  known  as  their  elasticity. 
We  can  conceive  of  this  regaining  of  dimensions  as  being  due  to  a  stress- 
resisting  force  inherent  in  the  material  and  which  we  can  designate  as  the 
force  of  elasticity,  yet  the  nature  of  the  force  is  unknown.    As  an  example, 
suppose  a  steel  rod  to  be  in  tension.     We  can  conceive  of  the  material 
particles  exerting  a  pull  upon  each  other.     This  we  call  stress.     And  we 
can  conceive  of  this  pull  being  resisted  by  a  force  inherent  in  each  particle. 
This  inherent  force  is  the  force  of  elasticity,  the  intensity  of  which,  in 
accordance  with  Newton's  Law,  is  necessarily  equal  to  the  stress  which  it 
resists. 

30.  Distortion. — In  this  book  the  total  change  of  form  of  a  body 
due  to  external  forces  will  be  known  as  distortion,  while  the  amount  of 
change  per  unit  of  dimension  of  a  body  will  be  known  as  unit  distortion. 
What  really  takes  place  in  every  case  is  cubic  distortion,  but,  in  deter- 
mining the  distortion  of  structures,  it  is  usually  necessary  to  consider  only 
the  distortion  of  the  length  of  their  parts,  and  hereafter  the  word  dis- 
tortion in  this  book  will  refer  only  to  the  distortion  of  length  unless  other- 
wise stated.     Suppose  a  steel  rod,  ten  feet  long,  is  suspended  from  one 
end  and  a  weight  is  hung  on  the  other  end.     If  the  weight  causes  the  rod 
to  stretch  one-tenth  of  an  inch  in  length,  the  distortion  of  its  length  or 
simple  distortion,  in  that  case,  is  one-tenth  of  an  inch,  while  the  unit 
distortion,  considering  an  inch  as  the  unit  of  length,  is   1/120   of  the 
distortion,  the  rod  being  ten  feet  long  or  120  inches. 

31.  Elastic  Limit. — Whenever  a  body  is  distorted  the  maximum 
amount  that  it  will  sustain   and  yet  return   to  its  original   form  when 
permitted  to  do  so,  it  is  said  to  be  distorted  to  the  elastic  limit.    If  a  body 
is  distorted  beyond  the  elastic  limit,  it  will  remain  shorter  or  longer  than 
its  original  length.     Then  we  say  that  the  body  has  taken  set.     If  a  body 
be  in  tension,  its  length  would  remain  longer,  and  if  in  compression  it 
would  remain  shorter. 

32.  Relation  of  Stress  and  Distortion. — Hooke  (in  1678)  was  the 
first  to  state  that  the  distortion  of  a  body  is  directly   proportional  to 
the  stress.     This  is  known  as   Hooke's   Law,  and  has  been  proven  by 
experience  to  be  practically  true,  provided  the  distortion  does  not  exceed 
the  elastic  limit.     Whenever  a  body  is  distorted  beyond  the  elastic  limit, 
the  distortion  increases  more  rapidly  than  the  stress. 


20  STRUCTURAL  ENGINEERING 

Suppose  a  steel  rod  having  a  length  L  and  a  uniform  cross-section 
A  is  distorted  an  amount  D  when  subjected  to  a  simple  stress  P;  then, 
according  to  Hooke's  Law,  if  the  stress  were  2P,  the  corresponding  dis- 
tortion of  its  length  would  be  2D;  if  3P,  it  would  be  3D;  and  so  on. 
While  this  direct  proportion  holds  good  in  all  cases  as  long  as  the  stresses 
are  within  the  elastic  limit,  yet  the  actual  value  of  D  in  any  case  will 
depend  upon  the  value  of  L  and  A,  being  directly  proportional  to  L,,  and 
inversely  proportional  to  A.  This  is  readily  seen,  for  it  is  obvious  that 
if  the  rod  were  two  feet  long  it  would  be  distorted  twice  as  much  when 
subjected  to  the  same  stress  as  a  rod  of  the  same  section  only  one  foot 
long,  since  each  foot  of  length  would  be  distorted  the  same  in  amount  in 
either  case. 

In  regard  to  the  cross-section,  it  is  obvious  that  if  the  area  is  two 
square  inches,  the  distortion  of  the  rod  will  be  only  one-half  as  much  as 
when  its  area  is  one  square  inch,  as  the  actual  stress  on  the  material  in  the 
rod  in  the  first  case  is  only  one-half  as  much  as  it  is  in  the  second  case. 
33.  Modulus  of  Elasticity  and  Determination  of  Simple  Dis- 
tortion. —  According  to  Hooke's  Law,  if  we  know  the  distortion  of  any  one 
piece  of  material  subjected  to  a  known  stress,  we  can  determine  the 
distortion  of  any  other  piece  of  the  same  kind  of  material  subjected  to  any 
known  stress  simply  by  direct  proportion. 

Suppose  it  is  found  by  experiment  that  a  rod  of  some  unknown 
material,  having  a  length  of  L  inches  and  a  uniform  cross-section  of  A 
square  inches,  is  distorted  D  inches  in  length  when  subjected  to  a  simple 
stress  of  P  pounds.  What  would  be  the  distortion  of  a  rod  of  the  same 
kind  of  material  having  a  length  of  L'  inches  and  a  uniform  cross-section 
of  A'  square  inches,  if  subjected  to  a  simple  stress  of  P'  pounds?  Let 
D'  be  the  distortion  required.  Reducing  everything  concerned  to  its 
lowest  terms,  which  is  done  for  convenience,  we  have 

The  unit  stress  in  the  first  rod        =  p  =  P/A, 
while  the  unit  distortion  in  inches  =  d  =  D/L. 
The  unit  stress  in  the  second  rod    =  pf  —  Pf/A'9 
while  the  unit  distortion  in  inches  =  d'  =  D'  /L'. 
By  direct  proportion  we  have 


Expressing  this  in  words,  we  say:  The  stress  (p)  per  square  inch 
in  the  first  rod  is  to  the  stress  (p')  per  square  inch  in  the  second  rod  as 
the  distortion  (d)  of  the  first  rod  per  inch  of  length  is  to  the  distortion 
(d')  of  the  second  rod  per  inch  of  length;  from  which  we  have  d'  =  dp'/pt 
which  is  the  distortion  of  each  inch  of  the  second  rod  ;  then,  of  course,  the 
total  distortion  of  the  rod  would  be  L'  times  this,  so  we  have 

D'  =  d'L'=-  p'L'  .          ..............................  (3). 

P 

Now  d/p  is  the  ratio  of  the  unit  distortion  to  the  unit  stress  given 
for  the  first  rod,  but  according  to  Hooke's  Law,  this  ratio  is  the  same  for 
any  piece  of  this  same  kind  of  material.  So  if  this  ratio  is  determined 
for  any  one  piece  of  this  material,  the  elastic  property  of  the  material  is 
known,  and  the  distortion  of  any  piece  of  the  material  can  then  be  deter- 
mined if  its  length,  area  of  cross-section,  and  stress  are  known. 


FUNDAMENTAL  ELEMENTS  21 

The  ratio  of  the  unit  distortion  to  the  unit  stress,  as  expressed  above, 
would  be  a  very  small  fraction  in  any  case,  so,  for  convenience,  we  can 
invert  the  expression  so  as  to  have  a  whole  number.  Then  we  have 
p/d  —  pf/d'  —  E,  which  is  a  constant  for  any  piece  of  the  same  kind  of 
material  composing  the  two  rods.  Now  substituting  1/-E  for  d/p  in 
equation  (2),  we  have  D'  =  p'L'/E.  E  would  be  known  as  the  Modulus 
of  Elasticity  of  the  material  composing  the  two  rods.  For  the  modulus  of 
elasticity  of  any  material  in  general  we  have 
_,  p  unit  stress 

Jl,  —~  —  -  -  -  y 

a      unit  distortion 

which  is  usually  designated  as  Young's  Modulus.  It  varies  with  the 
different  kinds  of  material,  but  is  practically  constant  for  all  pieces  of 
the  same  kind.  For  the  simple  distortion  of  any  piece  of  any  kind  of 
material  having  a  uniform  cross-section  and  being  subjected  to  a  simple 
stress,  we  have  the  general  equation 


where  D  =  distortion  of  the  piece  in  inches  (or  feet  if  L  be  taken  in  feet)  ; 
P  =  total  stress  on  the  cross-section  of  the  piece  in  pounds  ; 
A  =  area  of  cross-section  in  square  inches  ; 
L  =  total  length  of  piece  in  inches  or  feet; 
p  =  unit  stress  on  the  cross-section  of  the  piece; 
E  =  modulus  of  elasticity  of  the  material  composing  the  piece. 
The  modulus  of  elasticity  of  a  material  is  determined  by  experiment. 
The  average  values  of  the  moduli  of  elasticity  of  the  principal  materials 
used  in  structural  engineering  have  been  found  to  be  as  follows  : 
29,000,000  for  steel; 
26,000,000  for  wrought  iron; 
30,000,000  for  cast  steel; 
18,000,000  for  cast  iron; 

720,000   for  long-leaf  yellow  pine; 
555,000  for  white  oak. 

In  determining  the  modulus  of  elasticity,  it  is  laboratory  practice  to 
measure  the  distortion  in  inches,  in  which  case  the  length  of  the  test  piece 
must  always  be  reduced  to  inches;  but  in  using  Formula  (B),  L  may  be 
taken  either  in  feet  or  inches.  If  taken  in  feet,  the  distortion  will  be 
given  in  feet,  and  if  taken  in  inches,  it  will  be  given  in  inches. 

Problem  4.  Suppose  a  steel  rod  having  a  length  of  10  feet  and  a 
uniform  cross-section  of  2  square  inches  is  suspended  from  one  end  and 
supports  a  weight  of  30,000  pounds  hung  on  the  lower  end: 

(a)  What  will  be  the  distortion  of  the  rod  due  to  the  30,000  pounds  ? 
Referring    to    Formula    (B),    given    above,    we    have    in    this    case 

P  =  30,000;   p  =  30,000  *2  =  15,000;   E-  29,000,000.      Substituting   these 
values  in  the  formula,  we  have  for  the  distortion 

15,000x10  f        -t    .nch  (about) 

29,000,000 

(b)  What  will  be  the  distortion  of  the  rod  due  to  its  own  weight? 
In  that  case  the  stress  varies  from  zero  at  the  bottom  of  the  rod  to  a 

maximum  at  the  top.     For  convenience,  instead  of  using  the  numerals 


22  STRUCTURAL  ENGINEERING 

given  above,  let  L  be  the  length  of  the  rod  and  A  the  cross-section;  and 
let  w  be  the  weight  of  the  rod  per  foot  of  length.  Then  the  stress  on  the 
cross-section  at  a  point  b  (Fig.  14),  x  feet  from  the 
lower  end  of  the  rod,  will  be  P  =  wx,  which  is  really  the 
weight  of  the  rod  below  b. 

For  the  distortion  of  the  length  of  the  material  in 
the  rod  between  the  cross-section  at  b  and  the  cross- 
section  at  c,  which  is  an  infinitesimal  distance  dx  above 
b,  we  have 

wxdx 


which  corresponds  to  Formula   (B),  given  above. 

Now  as  x  is  a  variable,  varying  from  0  to  L,  the  last  expression  will 
give  the  distortion  of  any  infinitesimal  part  of  the  length  of  the  rod  any- 
where between  the  ends.  Then  the  total  distortion  of  the  rod  will  be  equal 
to  the  summation  of  the  distortion  of  these  parts ;  so,  for  the  total  distor- 
tion of  the  rod,  we  have 

wxdx      1   wL2 


/I 


AE    ~  2  AE 


Now  let  wL,  which  is  the  total  weight  of  the  rod,  be  represented  by  P. 
Then  substituting  in  (c)  we  have 


This  shows  that  the  distortion  of  the  rod  due  to  its  own  weight  is  one-half 
of  what  it  would  be  for  an  equal  weight  suspended  from  the  lower  end. 
The  weight  of  the  rod  =  6.8  X  10  =  68  Ibs.  Substituting  in  (d)  we  have 
for  the  distortion 


Problem  5.  What  will  be  the  distortion  of  length  of  a  solid,  circular, 
cast-iron  column  having  a  diameter  of  6  ins.  and  a  length  of  18  ft.,  when 
supporting  a  load  of  282,000  Ibs.? 

Problem  6.  What  will  be  the  distortion  of  an  eye-bar  8"  x  1  J"  x  39'-0" 
(c.c.  end  holes)  when  subjected  to  a  stress  of  16,000#n"? 

34.  Motion.  —  Strictly  speaking,  a  body  can  have  but  two  distinct 
kinds  of  motion,  one  known  as  translation  and  the  other  as  rotation. 
When  a  body  moves  as  a  whole  along  an  imaginary  line,  known  as  its 
path,  its  motion  is  translation,  particularly  in  reference  to  any  relatively 
fixed  position  in  the  immediate  vicinity  of  the  body;  but  if  it  either  moves 
around  or  turns  about  an  imaginary  line,  known  as  its  axis,  its  motion  is 
rotation  in  reference  to  that  line. 

There  is  one  fact  regarding  motion  which  should  never  be  overlooked, 
that  is,  motion  is  absolutely  relative,  and  that  the  kind  of  motion  is  always 
dependent  upon  the  position  of  reference.  For  instance,  the  motion  of  a 
particle  in  a  body  rotating  about  an  axis  is  rotation  in  reference  to  that 
axis;  yet  at  the  same  time  its  motion  (the  particles  in  the  axis  excepted) 
is  translation  when  referred  to  any  relatively  fixed  position  in  the  imme- 
diate vicinity  of  the  particle. 


FUNDAMENTAL  ELEMENTS 


23 


Fig.   15 


In  reference  to  time,  motion  is  either  uniform  or  variable;  in  refer- 
ence to  the  character  of  path  described  it  is  either  rectilinear  or  curvi- 
linear, being  rectilinear  when  the  path  is  a  straight  line  and  curvilinear 
when  a  curve. 

35.  Moment  of  a  Force. — The  moment  of  a  force  about  any  point 
is  a  measure  of  its  tendency  to  produce  rotation  of  the  body  upon  which 
it  acts  about  the  point  chosen  and  is  equal  to  the  perpendicular  distance 
from  the  point  to  the  line  of  action  of  the  force  multiplied  by  the  intensity 
of  the  force.     The  point  is  known  as  the  center  of  moment,  while  the 
perpendicular  distance  is  known  as  the  lever  arm  of  the  force,  or  simply 
"arm." 

In  the  case  shown  in  Fig.  15,  the  moments  of  the  forces  PI,  P2,  and 
P3  about  the  point  O  would  be  expressed  as  aPl,  bP2,  and  cP3,  respect- 
ively. As  is  readily  seen,  the  forces  PI 
and  P2  would  have  a  tendency  to  rotate 
the  body  A  clock-wise  about  0,  while  P3 
would  have  a  tendency  to  rotate  it  counter 
clock-wise.  Now,  taking  one  direction  as 
minus  and  the  other  direction  as  plus,  say, 
counter  clock-wise  minus,  and  letting  M 
be  the  algebraic  sum  of  the  moments  of 
the  three  forces  about  0,  we  have  M  = 
«P1  +  bP2  -  cP3,  which  is  the  equation 
of  moments  of  the  three  forces 
about  0. 

In  case  of  forces  not  in  the  same  plane,  the  moments  are  usually 
taken  about  a  line,  in  which  case  the  center  of  moment  of  each  force  is  the 
point  on  the  line  nearest  the  force.  The  tendency  of  rotation  about  the 
line  is  what  is  really  measured  in  that  case. 

It  is  obvious  that  the  moment  of  a  force  about  any  point  in  its  line 
of  action  is  zero,  for  in  that  case  its  arm  is  zero.  If  the  arm  is  taken  in 
feet  and  the  force  in  pounds,  the  moment  will  be  expressed  in  what  is 
known  as  foot  pounds,  but  if  the  arm  is  taken  in  inches  and  the  force  in 
pounds,  the  moment  will  be  expressed  in  what  is  known  as  inch  pounds. 

36.  A  Couple  and  Moment  of  Same.— Two  equal  and  opposite 
parallel  forces  acting  upon  a  body   (other  than  along  the  same  line  of 
action)   form  what  is  known  as  a  couple.     Thus  in  Fig.  16,  the  force  P 
applied  at  c  upon  the  body  AB  and  the  equal  and  opposite  force  P  applied 
at  b  form  a  couple. 

The  moment  of  a  couple  is  equal  to  the  distance  between  the  forces 

multiplied  by  the  intensity  of  one  of  the  $ 10 

forces.     For  example,  the  moment  of  the 

couple  shown  in  Fig.  16  is  M  =  Pd.  j« 

The  moment  of  a  couple  is  not  changed 
by  changing  the  center  of  moments,  that 
is,  the  moment  about  one  point  is  the  same    ft  I     \t> 
as  about  any  other  point  in  the  plane  of  tp 

the  forces.     This  is  readily  seen,  for  tak- 
ing moments  about  e  or  g,  we  have  M  =         ')U 
(es  +  sg)P  —  Pd;  taking  moments  about  s, 


iLJB 


Fig.   18 


24  STRUCTURAL  ENGINEERING 

we  have  M  =  P(se)  +  P(sg)  =  P(se  +  sg)  =  Pd;  and,  again,  taking 
moments  about  o,  we  have  M  =  P(ho)  -  P(ko)  =  P(ho  -  ko)  =  P(hk)  = 
Pd.  Now,  evidently,  if  the  moment  of  a  couple  is  the  same  for  all  points 
in  the  plane  of  the  forces,  a  couple  can  be  considered  to  be  applied 
anywhere  in  the  plane  of  its  forces,  which  means  that  we  can  consider 
any  couple  as  being  moved  bodily  to  any  desired  position  in  the  plane  of 
the  forces.  Of  course,  the  forces  must  act  upon  the  body  concerned. 

Any  two  couples  are  equivalent  when  their  moments  are  equal.  But 
the  forces  and  arms  in  the  two  cases  are  not  necessarily  equal.  Rotation 
or  tendency  of  rotation  is  always  due  to  the  action  of  a  couple.  No  single 
force  can  produce  rotation. 

37.  Concurrent  and  Non- Concurrent  Forces. — When  the  lines  of 

action  of  two  or  more  forces  acting  upon  a  body  meet  at  a  common  point, 
the  forces  are  said  to  be  concurrent,  while  if  their  lines  of  action  do  not 
so  meet,  the  forces  are  said  to  be  non-concurrent. 

38.  Conspiring  Forces. — When  the  lines  of  action  of  two  or  more 
forces  acting  upon  a  body  coincide,  the  forces  are  known  as  conspiring 
forces. 

39.  Resultant  and  Component  Forces   and   Graphical  Repre- 
sentation of  Same. —  A  body  acted  upon  by  two  or  more  forces  may, 
thereby,  have  a  tendency  to  move  in  more  than  one  direction  at  the  same 
instant ;  but  as  a  body  cannot  occupy  two  positions  at  the  same  time,  it  is 
evident  that  motion  can  take  place  in  but  one  direction,  which  is  known  as 
the  direction  of  the  resultant  of  the  two  or  more  forces.     Now,  it  is  evident 
that  a  single  force  acting  in  this  direction  with  a  certain  required  intensity 
could  produce. the  same  effect  as  that  produced  by  the  two  or  more  forces 
combined.     Such  a  force  would  be  known  as  the  resultant  of  the  two  or 
more  forces,  while  each  of  the  two  or  more  forces  would  be  known  as  a 
component  of  the  resultant  force. 

For   example,  if  two  concurrent   forces  PI   and  P2    (Fig.    17)    be 

applied  simultaneously  to  a  small 
body  resting  upon  a  smooth,  horizon- 
tal plane  at  A,  we  can  readily  see  that 
the  force  PI  would  have  a  tendency 
to  move  the  body  along  its  line  of 
action  Ac,  while  at  the  same  time  the 
force  P2  would  have  a  tendency  to 
move  the  body  along  its  own  line  of 
Fig.  17  action  Ab.  Now  we  know,  from 

reasons    given    above,   that    the    body 

cannot  move  along  both  of  the  lines  at  the  same  time,  and  we  can  readily 
see  that  it  would  not  move  along  either  of  the  lines,  as  the  two  tendencies 
of  motion  are  not  in  the  same  direction;  so,  undoubtedly,  if  the  two  forces 
moved  the  body,  the  motion  would  take  place  along  some  line  different 
from  either  of  the  two. 

Suppose  the  force  PI,  acting  alone  and  continuously  for  one  second, 
moved  the  body  from  A  to  B,  so  that  at  the  end  of  the  second  the  body 
would  be  at  B.  Then  suppose  the  force  PI  ceased  to  act  at  that  instant, 
and  suppose  the  body  stopped  instantly  at  that  point  and  the  force  P2 
were  then  applied  and  in  the  same  manner  moved  the  body  from  B  to  C  in 


FUNDAMENTAL  ELEMENTS  25 

one  second,,  so  that  at  the  end  of  the  next  second  the  body  would  be  at  C. 
Then,  the  two  forces  PI  and  P2,  each  acting  separately  for  one  second, 
would  have  moved  the  body  from  A  to  C  in  two  seconds,  but  if  they  had 
acted  simultaneously  upon  the  body  they  would  undoubtedly  have  accom- 
plished the  same  thing  in  one  second,  for  the  simple  reason  that  each 
force  would  have  been  acting  for  one  second,  as  in  the  case  when  acting 
separately,  and  with  the  same  intensity,  for  otherwise  they  would  lose 
their  identity.  By  this  it  is  not  meant  that  the  two  forces  acting  simul- 
taneously would  have  moved  the  body  along  the  same  paths  as  when 
acting  separately,  but  that  they  would  have  moved  it  from  A  to  C  along 
some  line  in  one  second.  We  cannot  conceive  of  these  forces  doing  any- 
thing  more  than  they  would  have  to  do  in  moving  the  body  from  A  to  C; 
then,  undoubtedly,  they  would  move  it  along  the  straight  line  AC  joining 
the  two  positions,  as  that  would  be  the  least  required  of  them  in  the 
transaction.  Then  the  line  AC  would  be  the  direction  of  the  resultant  of 
the  forces  PI  and  P2. 

Now  it  is  evident  that  the  two  forces  PI  and  P2  acting  simulta- 
neously upon  the  body  at  A  would  have  the  same  effect  as  a  single  force 
R  (Fig.  17)  would  have  if  acting  along  the  line  AC  with  such  an  intensity 
as  to  move  the  body  from  A  to  C  in  one  second.  Therefore,  the  two  forces 
PI  and  P2  could  be  replaced  by  this  single  force  R,  which  is  their 
resultant.  Each  of  the  forces  PI  and  P2  are,  in  that  case,  components 
of  the  resultant  force  R. 

Assuming  the  resistance  to  be  the  same  everywhere  on  the  plane — 
which  was  really  the  assumption  at  the  start — the  distance  AB  will  be 
directly  proportional  to  the  intensity  of  force  PI,  which  moved  the  body 
over  that  distance,  and,  likewise,  the  distance  BC  will  be  directly  propor- 
tional to  the  force  P2.  Then  as  the  sides  AB  and  BC  of  the  triangle 
ABC  (Fig.  17)  are  directly  proportional  to  the  forces  PI  and  P2, 
respectively,  it  is  evident  that  we  can  construct  a  similar  triangle  such 
that  the  sides  will  represent  to  convenient  scale  the  intensities  of  these 
forces.  Such  a  triangle  is  shown  in  Fig.  18,  where  the  side  PI,  parallel 
to  the  side  AB  in  Fig.  17,  is  drawn  to  represent  the  intensity  of  the  force 
PI  to  scale  (so  many  pounds  per  inch)  and  the  side  P2  parallel  to  the 
side  BC  in  Fig.  17  is  drawn  to  represent  the 
intensity  of  the  force  P2  to  the  same  scale  as  PI. 
Then  the  side  R,  which  will  be  parallel  to  the 
side  AC  in  Fig.  17,  will  represent  the  intensity 
of  the  resultant  force  R  to  the  same  scale  as  PI 
and  P2.  Now  it  is  evident  that  any  two  concur-  Fig.  18 

rent    forces    with   their    resultant    can   be   repre- 
sented in  magnitude  and  direction  by  the  sides  of  a  triangle  which  is 
known  as  a  Force  Triangle.     It  is  important  to  note  that  the  forces  act 
in  one  direction  around  the  triangle,  while  their  resultant  acts  in  the 
opposite  direction. 

Suppose,  instead  of  the  two  forces  PI  and  P2  acting  upon  the  small 
body  shown  in  Fig.  17,  that  there  be  four  concurrent  forces  PI,  P2,  P3, 
and  P4,  as  shown  in  Fig.  19.  The  resultant  R  of  the  two  forces  PI  and 
P2  is  determined  by  constructing  the  force  triangle  ABC  in  the  same 
manner  as  was  shown  above.  Then  as  R  is  the  resultant  of  the  two  forces 


26 


STRUCTURAL  ENGINEERING 


Fig.   19 


PI  and  P2,  it  will  replace  them,  and  the  remaining  forces  to  be  considered 
are  P3,  P4,  and  R.     Now  constructing  the  force  triangle  A  CD  (Fig.  19) 

for  the  force  R  and  P3,  we  have  their  re- 
sultant 121.  Next,  constructing  the  force 
triangle  ADE  for  the  two  remaining  forces, 
121  and  P4,  we  have  their  resultant,  R2, 
which  is  the  final  resultant,  or,  in  other 
words,  is  the  resultant  force  of  all  the 
forces  PI,  P2,  P3,  and  P4,  and  as  such  can 
replace  them.  We  now  have  a  polygon 
ABCDEA  wherein  the  forces  are  seen  to 
act  in  the  same  direction  around  the  poly- 
gon, while  their  final  resultant  122  acts  in 
the  opposite  direction.  Such  a  polygon  is 
known  as  a  Force  Polygon.  The  force 
polygon  ABCDEA  in  this  case  was  con- 
structed by  combining  force  triangles;  but 
it  is  evident  that  the  same  could  have  been 
constructed  by  laying  off  the  forces  PI,  P2,  P3,  and  P4  in  consecutive 
order,  thus  obtaining  the  part  ABCDE  of  the  polygon,  and  then  joining 
EA  for  completion  as  shown  in  Fig.  20. 

While  it  is  advisable  to  begin  with  some  force  as  PI  and  lay  the 
forces  off  in  consecutive  order  around  the  body  as  we  have  done  here,  yet 

it  is  not  absolutely  neces- 

£  sary  to  do  so,  for  it  will 

,E  be   seen   upon   inspection 

that  the  forces  may  be 
taken  in  any  order  (ex- 
cept that  they  must  be 
laid  off  so  that  they  act 
in  the  same  direction 
around  the  polygon)  as 
shown  in  Fig.  21,  where 
the  line  AE,  which  is  the 
final  resultant,  is  the 
same  in  length  and  direc- 
tion as  the  final  resultant 
AE  shown  in  Fig.  19  and 
Fig.  20. 

Now  it  is  evident  from  the  above  that  any  number  of  concurrent 
forces  with  their  resultant  can  be  represented  in  magnitude  and  direction 
by  the  sides  of  a  closed  polygon  known  as  a  force  polygon,  wherein  the 
forces  act  in  one  direction  around  the  polygon  while  their  resultant  acts 
in  the  opposite  direction,  the  direction  of  action  in  each  case  being  indi- 
cated by  the  arrow  points.  The  force  triangle  is  really  a  force  polygon 
wherein  the  component  forces  are  only  two  in  number. 

As  any  number  of  concurrent  forces  with  their  resultant  can  be 
represented  in  magnitude  and  direction  by  a  closed  polygon,  it  follows 
that  any  force  may  be  resolved  into  any  number  of  components,  the  only 
requirements  being  that  the  force  and  its  components  form  a  closed 
polygon.  Thus  we  can  construct  any  number  of  components  as  A  a,  ab, 


Fig.   20 


Fig.   21 


FUNDAMENTAL  ELEMENTS 


27 


be,  cd,  and  dB  for  the  force  AB,  as  shown  in  Fig.  22,  by  simply  drawing 
the  components  practically  at  will,  the  only  restrictions  being  that  they 
form  a  closed  polygon  with  the  force  AB.  Here  the  force  AB  is  consid- 
ered as  a  resultant,  which  is  obviously  permissible  in  any  case.  We  would 
say  that  the  force  AB  was  resolved  into  its  component  forces  Aa,  ab,  be, 
cd,  and  dB.  Resolving  forces  into  components  is  known  in  mechanics  as 
the  Resolution  of  Forces.  The  most  convenient  components  into  which  a 
force  can  be  resolved  are  two  components  whose  directions  are  at  right 
b 


Fig.   22 


Fig.   23 


B 


angles  to  each  other,  and  are  known  as  rectangular  components.  Thus 
AD  and  DB  or  AE  and  EB  (Fig.  23)  are  rectangular  components  of  the 
force  AB.  It  is  obvious  that  a  force  can  be  resolved  into  any  number  of 
pairs  of  rectangular  components — or  any  number  of  any  other  kind. 

It  will  now  be  shown  that  the  above  is  as  true  for  non-concurrent 
forces  as  it  is  for  concurrent  forces.  Let  PI,  P2,  and  P3  (Fig.  24) 
represent  three  non-con- 
current  forces  acting  »s\  V^L, 
upon  the  body  AB.  As  pX\  \ 
far  as  motion  or  ten-  ^^i 
dency  of  motion  are  con- 
cerned, any  of  these 
forces  may  be  considered 
as  applied  at  any  point 
along  their  line  of  action ; 
however,  in  making  such 
assumptions  in  any  case, 
we  are  compelled  to  con- 
ceive that  there  always 
exists  the  necessary  ma- 
terialistic connection  be- 
tween such  points  of  ap- 
plication and  the  body 
concerned.  Then,  if  we 
prolong  the  lines  of  action  of  the  two  forces  PI  and  P2  until  they  inter- 
sect at  0,  we  can  consider  them  as  being  two  concurrent  forces  applied 
at  0,  and  by  constructing  the  force  triangle  OCD  (where  OC  =  P2  and 
CD  -  PI  to  scale)  we  have  their  resultant  given  in  amount  and  direction 
by  the  line  OD.  Now  this  resultant,  which  we  will  designate  as  121,  can  be 
considered  as  being  applied  at  any  point  along  yy,  its  line  of  action  (the 
same  as  any  other  force).  Then  prolonging  the  line  of  action  yy  of  this 
resultant  121  until  it  intersects  the  line  of  action  of  the  third  force  P3  at 
0',  and  constructing  the  force  triangle  O'EF,  we  have  the  resultant  122, 
which  is  the  final  resultant,  or,  in  other  words,  the  resultant  of  all  three 


Fig.  24 


28  STRUCTURAL  ENGINEERING 

forces,  PI,  P2,  and  P3.  This  final  resultant  R2  can  be  considered  as 
being  applied  at  any  point  upon  its  line  of  action  xx. 

From  the  above  it  is  evident  that  the  resultant  of  any  number  of 
intersecting  non-concurrent  forces  can  be  represented  graphically  by  first 
selecting  any  two  forces  and  prolonging  their  lines  of  action  until  they 
intersect;  then  constructing  a  force  triangle  and  obtaining  the  resultant 
of  these  two  forces ;  then  prolonging  the  line  of  action  of  this  resultant 
until  it  intersects  with  the  line  of  action  of  a  third  force;  and  then  again 
constructing  a  force  triangle  and  obtaining  the  resultant  of  the  first 
resultant  and  the  third  force;  and  again  prolonging  the  line  of  action  of 
this  second  resultant  until  it  intersects  the  line  of  action  of  a  fourth  force ; 
and  so  on  until  all  of  the  forces  are  combined.  The  last  resultant  will  be 
the  resultant  of  all  of  the  forces,  or  the  final  resultant. 

As  an  additional  example,  let  PI,  P2,  P3,  P4,  and  P5  represent  five 
non-concurrent  forces  acting  upon  the  body  AB,  as  shown  in  Fig.  25  (a). 
Let  us  first  take  the  two  forces  PI  and  P2  and  prolong  their  lines  of  action 
until  they  intersect  at  0.  Then  by  constructing  the  force  triangle  ODC 
we  obtain  their  resultant  JR1.  Then  by  prolonging  the  line  of  action  of 
this  resultant  until  it  intersects  the  line  of  action  of  a  third  force  P3  at  0', 
and  constructing  the  force  triangle  O'EF,  we  obtain  R2,  the  resultant  of 
Rl  and  P3.  Then  prolonging  the  line  of  action  of  R2  until  it  intersects 
the  line  of  action  of  a  fourth  force  P4  at  0",  and  constructing  the  force 


(a) 

Fig.  25 

triangle  0"GH,  we  obtain  123,  the  resultant  of  R2  and  P4.  Then  pro- 
longing the  line  of  action  of  R3  until  it  intersects  the  line  of  action  of  a 
fifth  force  P5  at  0" ',  and  constructing  the  force  triangle  0"  '  KL,  we 
obtain  the  resultant  .R4,  which  is  the  final  resultant,  or,  in  other  words, 
the  resultant  of  all  the  forces  PI  ...  P5. 

So  far  our  treatment  of  non-concurrent  forces  does  not  seem  to  have 
much  in  common  with  our  treatment  of  concurrent  forces,  but  let  us  go  a 
little  further  and  see  what  we  can  observe.  The  force  triangles  ODC, 
O'EF,  0"GH,  and  O" '  KL  are  the  same  in  kind  as  we  found  in  our 
treatment  of  concurrent  forces.  Now  suppose  the  force  triangle  ODC  in 
Fig.  25  (a)  be  moved  bodily  and  parallel  to  itself  to  the  position  ODC  in 
Fig.  25  (6).  This  would  not  change  the  triangle  in  the  least.  Next 
suppose  the  force  triangle  O'EF  be  moved  in  the  same  manner  to  Fig.  25 
(6),  taking  the  position  OCFf  the  vertex  0'  coinciding  with  0  in  Fig. 
25  (6)  and  E  with  C.  Next,  suppose  the  force  triangles  0"GH  and 
0" ' KL  (in  Fig.  25  (a))  to  be  moved  in  the  same  manner  to  the  position 
in  Fig.  25  (6)  so  that  0"  and  0" '  would  fall  at  O.  Then  these  two 
triangles  would  take  the  positions  OFK  and  OKL,  respectively.  By  thus 
grouping  the  force  triangles  we  have  constructed  the  force  polygon 


FUNDAMENTAL  ELEMENTS  29 

ODCFKLO,  which  is  the  same  kind  of  a  force  polygon  as  was  constructed 
for  concurrent  forces  in  Fig.  19,  and  it  can  be  drawn  in  the  same  manner. 
This  shows  that  any  number  of  non-concurrent  forces  with  their  resultant, 
the  same  as  concurrent  forces,  forms  a  closed  polygon,  known  as  a  force 
polygon,  and  hence  the  resultant  of  any  number  of  any  kind  of  external 
forces  acting  upon  any  body  whatever  can  be  determined  simply  by  con- 
structing a  force  polygon  representing  each  in  intensity  and  direction,  at 
the  same  time  placing  the  forces  so  that  they  act  in  the  same  direction 
around  the  polygon. 

As  a  general  case,  let  Fig.  26  represent  a  body  acted  upon  by  four 
non-concurrent  forces  PI  .  .  .  P4.  We  can  determine  the  resultant  of 
these  four  forces  by  simply  taking  any  one  of  them,  say,  P4,  and  drawing 
a  line  as  AB  equal  and  parallel  to  it.  Then  from  B  draw  BC  equal  and 
parallel  to  P3.  Then  from  C  draw  CD  equal  and  parallel  to  P2.  Then 
from  D  draw  DE  equal  and  parallel  to  PI. 
Then  joining  E  and  A  we  have  EA,  which  /r,/ 
represents  the  resultant  of  the  four  forces  in  \J  IRsV^ 

intensity  and  in  direction.   This  same  method      JL — *• j— j 

of  procedure  holds  for  both  concurrent  and 

non-concurrent  and  also  for  conspiring  forces. 
The  conspiring  forces  would  simply  be  laid 
off  in  one  straight  line.  Their  resultant  would 
be  represented  by  the  distance  from  the  last 
force  laid  off  to  the  starting  point. 

40.     Analytical  Representation  of  Resultant  and  Component 
Forces. — From  the  force  triangle  shown  in  Fig.   18  (Art.  39)  -we  have 

R2  =  Pl2  +  P22  +  2  cos  <f>  PlxP2 (1), 

where  </>  is  the  angle  which  the  lines  of  action  of  the  forces  PI  and  P2 
make  with  each  other.  Any  one  of  the  forces  PI,  P2,  R,  and  the  angle  <f> 
can  be  determined  from  the  above  equation,  provided  the  other  three  are 
given.  When  <£  =  90°,  the  above  equation  becomes 


which  is  the  "rectangular  equation"  that  results  whenever  a  force  is 
resolved  into  rectangular  components.  As  a  rule,  it  is  much  more  con- 
venient to  express  these  rectangular  components  in  terms  of  the  force 
and  angles  which  they  make  with  the  force.  Thus,  from  Fig.  27,  we  have 

12  cos  0  =  PI;  12  =  —  =Plxsec0; 

Rcos6  =  P2;  cos</> 

Rsind>  =  P2',  P2 


The  most  useful  thing  to  be  observed  from  the  above  is  that  the 
rectangular  component  of  a  force  along  any  line  is  equal  to  the  force 
multiplied  by  the  cosine  of  the  angle  which  the  force  makes  with  the  line. 

In  the  'case  of  conspiring  forces,  it  is  evident  that  the  resultant  of 
two  or  more  conspiring  forces  is  equal  to  their  algebraic  sum,  for  we  can 
consider  the  forces  acting  in  one  direction  as  plus  and  those  acting  in  the 
opposite  direction  as  minus.  Then  by  adding  all  of  the  plus  forces 
together  and  all  of  the  minus  forces  together  and  subtracting  the  lesser 


30 


STRUCTURAL  ENGINEERING 


from  the  greater  sum,  we  shall  thus  obtain  the  intensity  and  direction  of 

action  of  their  resultant. 

The  resultant  of  three  or  more  forces 
can  be  determined  analytically  most  readily 
by  resolving  the  forces  into  their  horizontal 
and  vertical  components,  thus  obtaining  a 
rectangular  equation  where  the  square  of 
the  algebraic  sum  of  the  horizontal  com- 
ponents plus  the  square  of  the  algebraic 
sum  of  the  vertical  components  is  equal  to 

the  square  of  the  resultant.     Thus  from  Fig.  19  (Art.  39)  we  have 


Fig.   27 


But  aE  =  ab  +  be  4  cE  =  P2cos6l  +  P3cos0,  +  P4cos03, 

which   is   the   algebraic   sum   of   the   vertical   components   of   the    forces 
PI,  P2,  P3,  andP4; 


and 


=  AB  +  Bd-Cf-Dc  =  PI  +  P2sm01  -  P3sin02  -  P4sin<93, 


which  is  the  algebraic  sum  of  the  horizontal  components  of  the  same 
forces,  as  is  readily  seen  from  Fig.  19.  613  62,  etc.,  are  the  angles  which 
the  forces  make  with  the  vertical  line  aE.  PI  has  no  vertical  component, 
as  the  cosines  of  the  angles  it  makes  with  the  vertical  is  O  —  the  angles 
being  90°  —  and,  the  sine  of  90°  being  1,  it  is  evident  that  its  horizontal 
component  is  equal  to  PI. 

41.     Proposition.  —  The   moment   of   the   resultant   of   two   or  more 
forces  about  any  point  in  the  plane  of  the 
forces  is  equal  to  the  algebraic  sum  of  the 
moments  of  the  'two  or  more  forces  them- 
selves, about  the  same  point. 

As  the  resultant  of  two  or  more  forces 
is  a  force  which  will  produce  the  same  effect 
(as  far  as  motion  or  tendency  of  motion  are 
concerned)  as  the  two  or  more  forces  com- 
bined, it  is  practically  self-evident  that  the 
above  proposition  is  true.  However,  it  is 
seen  to  be  true  from  the  following  demon- 
stration  : 

Let  R  (Fig.  28)  represent  the  resultant  of  two  concurrent  forces  PI 
and  P2,  acting  upon  a  body  at  A,  and  let  the  intensity  of  these  forces  and 
their  resultant  be  represented  by  the  sides  of  the  force  triangle  ABC. 
Select  any  point  0  as  the  center  of  moments.  Then  we  are  to  prove  that 
the  moment  of  R  about  O  is  equal  to  the  algebraic  sum  of  the  moments  of 
the  component  forces  PI  and  P2  about  the  same  point.  Now,  it  makes  no 
difference  where  0  is  taken,  we  can  always  draw  a  line  through  0  parallel 
to  the  line  of  action  of  the  resultant  R.  Then  in  this  case  draw  the  line 
OO'  parallel  to  R  and  from  A  draw  h'  perpendicular  to  the  line  00'; 
then  the  moment  of  the  resultant  R  about  O'  is  the  same  as  it  is  about  O, 
since  h'  =  h. 

For  the  equation  of  moments  about  0',  if  the  above  proposition  be 
true,  we  have  Rh'  =  a'Pl  +  d'P2;  but  a'  =  h'cos<j>  and  d'  =  h'cosO,  and 
substituting  we  have  Rh'  =  (Plcos<j>-\-P2cosO)h't  or  R  =  Plcos<j>  +  P2eos6. 


Fig.  2s 


FUNDAMENTAL  ELEMENTS  31 

But  from  the  force  triangle  ABC  we  have  R  =  Plcos  <J>  +  P2cos  6;  then 
substituting  this  value  for  R  in  the  last  equation,  we  have  (Plcos<£-f 
P2cos#)  —  (Plcos<£  +  P2cos#)  an  identity,,  which  proves  the  proposition 
when  0'  is  the  center  of  moments.  Now,,  as  the  moment  of  R  about  0  is 
the  same  as  it  is  about  0'  ,  it  remains  for  us  to  prove  that  the  change  in 
the  algebraic  sum  of  the  moments  of  PI  and  P2,  due  to  the  changing  of 
the  center  of  moments  from  0'  to  0  is  zero.  That  is,  we  are  to  prove  that 
(a  -  a')Pl  -  (d'  -d)P2  =  0,  or  (a  -  a')Pl  =  (df  -  d)P2.  Now  from  Fig.  28 
we  have  a  —  af  —  c  sin  <£  and  d'  -d  =  c  sin  6.  Then  substituting  these 
values,  and  cancelling  c,  we  have 


which  is  readily  seen  to  be  true  from  the  force  triangle  ABC,  where  we 
have 


So  we  have  thus  proven  our  proposition  for  the  case  of  two  forces.  Now 
as  it  was  shown  in  Art.  39  that  the  force  polygon  representing  any 
number  of  forces  was  really  made  up  by  combining  force  triangles,  it  is 
obvious  that  we  could  show  that  the  moment  of  the  final  resultant  of  any 
number  of  forces  about  any  point  is  equal  to  the  algebraic  sum  of  the 
moments  of  the  forces  about  the  same  point  by  simply  considering  the 
resultants  and  components  in  each  triangle  in  consecutive  order. 

42.  Condition  of  Equilibrium.  —  We  say  a  body  is  in  equilibrium 
whenever  it  is  either  at  rest  or  in  uniform  motion.  The  only  state  of 
equilibrium  here  considered  shall  be  that  of  rest,  as  uniform  motion  is  of 
little  interest  to  us  in  structural  mechanics.  As  there  are  but  two  kinds 
of  motion,  translation  and  rotation,  it  is  evident  that  a  body  will  be  at  rest 
if  neither  of  these  motions  takes  place,  and  that  the  problem  regarding 
equilibrium  thus  reduces  to  the  investigation  of  the  conditions  conducive 
to  these  two  motions. 

Our  only  conception  of  equilibrium  of  forces  is  that  of  two  equal  and 
opposite  conspiring  forces  balancing  each  other  through  the  body  upon 
which  they  act.  Then,  in  order  that  a  body  be  in  equilibrium,  it  is  obvious 
that  the  forces  acting  upon  it  must  reduce  to  that  condition.  The  forces 
themselves,  or  resultants,  or  a  combination  of  the  forces  and  resultants, 
may  form  a  pair  or  several  pairs  of  equal  and  opposite  conspiring  forces, 
so  that  all  of  the  forces  acting  upon  a  body  will  thus  be  balanced,  or,  as 
we  say,  be  in  equilibrium,  and  such  being  the  case,  the  body  upon  which 
they  act  will  be  in  equilibrium.  It  makes  no  difference  what  combination 
we  choose  to  consider  the  forces  forming,  for  so  long  as  we  can  show  that 
the  whole  system  of  forces  acting  upon  a  body  reduces  to  equal  and 
opposite  conspiring  forces,  we  are  assured  that  the  body  upon  which  the 
system  acts  is  in  equilibrium,  and  of  course  the  converse  will  be  true. 

Referring  to  Fig.  17  (Art.  39),  it  is  obvious  that  if  a  force  equal  in 
intensity  to  the  resultant  of  the  forces  PI  and  P2  be  applied  to  the  body 
at  A  so  that  it  would  act  along  the  line  AC  in  the  direction  from  C  to  A, 
it  would  balance  the  forces  PI  and  P2  and  the  body  would  thus  be  in 
equilibrium  under  the  action  of  these  three  forces.  Here  it  is  seen  that 
the  resultant  of  the  two  forces  PI  and  P2  and  the  balancing  force  would 
be  two  equal  and  opposite  conspiring  forces  to  which  the  system  reduces 
when  the  body  is  in  equilibrium. 


32  STRUCTURAL  ENGINEERING 

And,  referring  to  Fig.  19  (Art.  39),  it  is  obvious  that  if  a  force  equal 
in  intensity  to  the  final  resultant  R2  be  applied  to  the  body  at  A  so  that 
it  would  act  along  the  line  EA  in  the  direction  from  E  to  A,  it  would 
balance  all  of  the  forces  PI,  P2,  P3,  and  P4,  and  thus  the  body  would  be 
in  equilibrium  under  the  action  of  the  five  forces  PI,  P2,  P3,  P4,  and  the 
balancing  force.  Here  it  is  seen  that  the  resultant  of  the  forces  PI,  P2, 
P3,  and  P4  and  the  balancing  force  are  two  equal  and  opposite  conspiring 
forces  to  which  the  system  reduces  when  the  body  is  in  equilibrium. 

Further,  referring  to  Fig.  25  (a)  (Art.  39),  it  is  obvious  that  if  a 
force  equal  and  opposite  to  R4:  be  applied  along  the  line  S'O"  '  it  would 
balance  all  of  the  forces  PI  .  .  .  P5,  acting  upon  the  body  AB,  and  thus 
the  body  would  be  in  equilibrium.  Here  again  it  is  seen  that  the  system 
reduces  to  equal  and  opposite  conspiring  forces  when  equilibrium  exists. 
So  it  is  in  all  cases. 

It  will  be  observed  from  the  above  that  any  force  acting  upon  a  body 
in  equilibrium  will  always  form  an  equal  and  opposite  conspiring  force 
with  the  resultant  of  all  the  other  forces  acting  upon  the  body. 

Beyond  a  question,  a  body  will  move  in  translation  if  the  forces 
acting  upon  it  have  a  resultant.  Then,  evidently,  a  body  will  be  in 
equilibrium,  as  far  as  translation  is  concerned,  whenever  the  forces  acting 
upon  it  have  no  resultant.  Then  as  the  forces  acting  upon  a  body  in 
equilibrium  must  have  no  resultant — in  order  that  no  translation  may 
take  place — it  is  obvious  that  the  force  polygon  representing  them,  instead 
of  having  a  final  resultant,  as  R  2  in  Fig.  19,  will  simply  extend  on  around 
and  close  on  the  starting  point  without  a  resultant.  This  would  give  us  a 
force  polygon  wherein  all  the  forces  are  indicated  to  act  in  the  same 
direction  around  the  polygon.  From  this  the  following  practical  state- 
ment can  be  made:  A  body  will  be  in  equilibrium  as  far  as  translation  is 
concerned  whenever  the  forces  acting  upon  it  (when  represented  graph- 
ically to  scale)  will  form  a  closed  polygon  wherein  all  of  the  forces  act 
in  the  same  direction  around  the  polygon,  or,  in  other  words,  the  force 
polygon  will  close  without  a  resultant,  and  the  converse  is  just  as  true, 
that  is,  if  a  body  be  in  equilibrium  the  polygon  will  close.  This  is  true 
regardless  of  whether  the  forces  be  non-concurrent,  conspiring,  or  con- 
current forces,  for  it  is  evident  that  a  body  will  have  motion  of  translation 
if  a  resultant  exists,  as  the  resultant  is  equivalent  to  a  single  force,  in 
which  case  motion  is  inevitable;  and  if  no  resultant  exists,  motion  of 
translation  will  not  take  place,  as  there  would  be  no  force  to  produce  it. 
All  this  is  manifestly  true  regardless  of  condition. 

As  the  resultant  of  concurrent  or  conspiring  forces  always  passes 
through  a  common  point  of  action,  and  the  resultant  of  all  the  forces 
except  one  always  forms  an  opposite  and  equal  conspiring  force  with  the 
remaining  one,  in  case  of  equilibrium,  and  the  moments  about  any  point 
of  these  two  equal  and  opposite  conspiring  forces  are  equal  and  have 
opposite  signs,  it  is  evident  that  rotation  will  not  take  place  if  the 
equilibrium  polygon  closes.  So  in  the  case  of  concurrent  or  conspiring 
forces,  if  the  force  polygon  closes,  we  need  go  no  further  in  our  investi- 
gations regarding  equilibrium,  for  if  no  resultant  exists,  no  motion  of 
rotation,  as  well  as  no  motion  of  translation,  will  take  place. 

But  in  regard  to  non-concurrent  forces  the  case  is  different,  for  here 
the  forces  may  reduce  to  two  equal  and  opposite  forces,  which  indicates 


FUNDAMENTAL  ELEMENTS  33 

that  no  resultant  exists,  and  thus  the  force  polygon  would  close,  indicating 
no  motion  of  translation ;  but  at  the  same  time  the  two  forces  might  not  be 
conspiring  forces,  in  which  case,  while  motion  of  translation  would  not 
take  place,  motion  of  rotation  would,  as  the  two  forces  would  form  a 
couple  which  would  produce  motion  of  rotation.  For  example,  the  forces 
acting  upon  the  body  AB  shown  at  (a)  (Fig.  29)  could  form  a  closed 
polygon  as  shown  at  (fr),  which  indicates 
that  no  resultant  exists,  and  such  being  ~\  /P2 

the  case,  no  translation  would  take  place ;        »  pi — * -. 

but  by  mere  inspection  of  the  figure  at  $J  i — ' 

(a)    we   can   see   that   the    forces   would  (®)     '      \P4 

form   a   couple   which   would   rotate   the 

body  AB   counter  clock-wise   and  hence  Fig.  29 

equilibrium  would  not  exist.     So  to  state 

the  condition  of  equilibrium  fully,  we  say :  A  body  is  in  equilibrium  when 

both  the  resultant  and  the  algebraic  sum  of  the  moments  about  every  point 

of  the  external  forces  acting  upon  it,  equal  zero. 

In  addition  to  the  above  statement,  we  may  add  that  a  body  is  in 
equilibrium,  as  far  as  motion  of  translation  is  concerned,  when  the 
algebraic  sum  of  both  the  horizontal  and  vertical  components  of  the  forces 
acting  upon  it  is  equal  to  zero,  for  it  was  shown  in  Art.  40  that  the  sum 
of  the  squares  of  the  horizontal  and  vertical  components  of  any  number 
of  forces  is  equal  to  the  square  of  their  resultant,  and,  of  course,  if  the 
components  are  equal  to  zero,  the  resultant  will  be  equal  to  zero.  Further, 
we  can  state  that  the  sum  of  the  components  of  the  forces  acting  upon  a 
body  in  equilibrium  is  zero  along  any  line  whatsoever,  for  it  is  evident 
that  if  a  component  exists  along  any  line,  a  resultant  would  exist  in  some 
direction,  and  hence  motion  of  translation  would  surely  take  place  in  that 
direction. 

43.    Point  of  Application  of  a  Force  in  Relation  to  Stress  and 

Equilibrium. — In  order  to  determine  the  nature  of  the  stress  produced  in 
any  case  it  is  absolutely  necessary  to  know  the  actual  point  of  application 
of  the  force  producing  the  stress.  Each  force  in  that  case  is  indicated  to 
act  at  some  particular  point  as  shown  in  Art.  26,  while  in  reference  to 
equilibrium  (as  stated  in  Art.  39),  a  force  may  be  considered  as  being 
applied  at  any  point  along  its  line  of  action,  in  which  case,  of  course,  we 
are  compelled  to  conceive  of  there  always  being  the  necessary  materialistic 
connection  between  the  point  of  application  and  the  body  concerned. 

In  the  case  of  simple  stress,  whenever  a  force  acts  toward  a  body, 
the  stress  produced  in  the  body  by  the  force  will  always  be  compression, 
but  if  a  force  acts  away  from  a  body,  the  stress  will  be  tension. 

As  a  simple  case,  let  AB  (Fig.  30)  represent  a  steel  rod  acted  upon 
by  two  equal  and  opposite  conspiring  forces,  PI  and  P2.  If  PI  be 
applied  at  a  and  P2  at  b,  as  indicated,  it  is  obvious  that  the  rod  would  be 
in  tension.  Now  let  us  suppose  the  forces  interchanged,  PI  being  applied 
at  b  and  P2  at  a,  then  it  is  obvious  that  the  rod  would  be  in  compression. 
Thus,  we  see  that  by  changing  the  point  of  application  of  the  forces,  the 
stress  in  the  rod  would  be  completely  reversed,  while  the  state  of 
equilibrium  would  not  be  disturbed. 

Then  again,  suppose  PI  be  applied  at  a  and  P2  at  c,  then  it  is 
obvious  that  the  stress  in  the  rod  between  a. and  c  would  be  the  same  as 


34  STRUCTURAL  ENGINEERING 

when  P2  is  applied  at  b  and  PI  at  a,  while  the  stress  in  the  rod  between 
c  and  b  would  be  zero.  So  it  is  readily  seen  that  any  change  in  the  point 
of  application  of  either  of  the  forces  will  produce  a  corresponding  change 


Fig.   30 

upon  the  rod  in  reference  to  the  stress,  but  it  is  just  as  readily  seen  that 
any  such  change  will  riot  disturb  the  state  of  equilibrium,  for  it  is  evident 
that  the  forces  PI  and  P2  will  balance  each  other  so  long  as  their  points 
of  application  remain  in  the  body  anywhere  on  their  common  line  of  action 
between  the  points  a  and  b.  As  for  that,  either  of  the  forces  can  even  be 
considered  as  being  applied  at  any  point,  as  o,  off  of  the  body  altogether, 
in  which  case,  of  course,  we  would  have  to  consider  the  rod  as  extending 
to  o  or  being  rigidly  connected  to  it  in  some  way. 

As  another  case,  let  three  non-concurrent  forces,  PI,  P2,  and  P3 
(Fig.  31)  be  applied  to  the  body  ZZ  at  the  points  a,  b,  and  c,  respectively. 
Now,  as  far  as  equilibrium  is  concerned,  we  can 
consider  any  of  these  forces  as  being  applied  at 
any  point  along  their  line  of  action,  so  prolong 
the  lines  of  action  of  the  two  forces  PI  and  P2 
(see  Art.  39)  until  they  meet  at  A.  Then  we  can 
consider  these  two  forces  as  two  concurrent  forces 
applied  at  A,  and  constructing  the  force  triangle 
ABC,  we  have  the  intensity  and  direction  of  their 
resultant  Rl,  which  can  be  considered  as  a  force 
being  applied  at  any  point  along  its  line  of  action 
Cy.  Then  prolong  the  line  of  action  of  the  force 
P3  until  it  meets  the  line  Cy  at  D.  Then  we  can 
rig.  31  consider  that  Rl  and  P3  are  two  concurrent 

forces  applied  at  D.     Then  constructing  the  force 

triangle  DEF,  we  have  their  resultant  R2,  which  is  the  final  resultant  of 
the  three  forces  PI,  P2,  and  P3. 

Now  it  is  evident  that  if  a  force  equal  to  Rl  be  applied  to  the  body 
along  the  line  Cy  it  would  produce  the  same  motion  as  PI  and  P2  if  it 
acted  in  the  same  direction  as  indicated  by  Rl,  but  if  it  acted  in  the 
opposite  direction  it  would  balance  them.  Then  again,  it  is  evident  that  if 
a  force  equal  to  the  resultant  R2  be  applied  to  the  body  along  the  line  Dx 
it  would  produce  the  same  motion  of  translation  as  the  forces  PI,  P2,  and 
P3  combined,  providing  it  acted  along  the  line  Dx,  in  the  same  direction 
as  indicated  by  R2;  but  if  it  acted  in  the  opposite  direction,  it  would 
balance  the  three  forces  PI,  P2,  and  P3. 

Now  in  regard  to  stresses,  it  is  obvious  that  the  stress  produced  upon 
the  body  ZZ  by  the  action  of  the  three  forces  PI,  P2  and  P3  will  be  of  a 
very  complicated  nature.  We  can  readily  see  that  a  force  equal  to  the 
resultant  121,  if  applied  at  o  or  at  any  other  point  in  or  on  the  body  along 
the  line  Cy,  would  not  produce  the  same  stress  as  the  two  forces  them- 
selves produce.  And  it  is  more  readily  seen  that  a  force  equal  to  the 
resultant  R2  applied  along  the  line  Dx  would  not  produce  the  same  stress 
in  the  body  as  the  forces  PI,  P2,  and  P3  actually  produce.  So  it  is 


FUNDAMENTAL  ELEMENTS  35 

evident  that  a  force  representing  the  resultant  of  two  or  more  forces, 
while  it  will  produce  the  same  effect  as  the  forces  themselves  as  far  as 
motion  or  tendency  of  motion  is  concerned,  yet  will  not  necessarily  produce 
the  same  effect  in  reference  to  stress.  However,  in  the  case  of  concurrent 
forces,  we  assume  the  stress  produced  by  a  force  representing  a  resultant 
to  be  the  same  as  that  produced  by  the  components,  especially  when  the 
forces  meet  at  the  surface  of  the  body  considered.  Some  cases  are  very 
complicated  and  require  very  careful  consideration,  but  we  can  always 
obtain  a  reasonable  approximation  which  will  come  within  the  limits  of 
practicability,  and  with  this  we  must  be  contented. 

44.  Equilibrium  of  Couples.— In  regard  to  equilibrium  of  couples, 
there  is  one  important  fact  that  should  always  be  borne  in  mind,  and  that 
is,  it  always  requires  another  couple  to  balance  a  couple,  and  the  couples 
must  be  equivalent  to  each  other  and  act  in  opposite  direction  to  each 
other.    As  to  the  proof  that  it  requires  another  couple  to  balance  a  couple, 
we  have  the  following: 

We  know  from  the  definition  that  a  couple  can  have  no  single 
resultant,  as  the  resultant  always  reduces  to  zero,  and  therefore  would 
not  balance  a  single  force,  and  hence  any  single  force  or  any  system  of 
forces  reducing  to  a  single  resultant  could  not  balance  a  couple.  Then, 
evidently,  the  forces  balancing  a  couple  must  have  a  resultant  equal  to 
zero  and  an  equal  and  opposite  moment,  which  would  undoubtedly  con- 
stitute another  couple. 

45.  Graphical  Determination  of  the  Resultant  of  Parallel  Forces 
or  Those  Nearly  Parallel. — First  Method:    In  the  case  of  intersecting 
forces,  the  "final  resultant  can  be  fully  determined  in  its  true  position  by 
prolonging  in  consecutive  order  the  lines  of  action  of  both  the  forces  and 
their  successive  resultants  to  intersection,  and  constructing  a  force  triangle 
at  each  intersection,  as  shown  in  Art.  39,  but  it  is  obvious  that  in  the  case 
of  parallel  forces  or  forces  nearly  parallel  the  same  method  will  not  apply 
as  such  forces  either  do  not  intersect  at  all  or  their  intersection  is  beyond 
practical  limits.     However,  by  resorting  to  the  resolution  of  forces,  the 
resultant  of  parallel  forces  and  those  nearly  parallel  can  be  determined 
as  readily  in  its  true  position  as  the  resultant  of  intersecting  forces.     As 
an  example,  let  AB  at  (a)  in  Fig.  32  represent  a  body  acted  upon  by 
three  non-concurrent  forces  PI,  P2,  and  P3,  which  are  nearly  parallel. 
Select  some  convenient  point  on  the  line  of  action  of  one  of  the  forces, 
say,  point  0  on  the  line  of  action  of  PI.     Then  at  this  point  0  resolve  PI 
into  any  two  component  forces  as  c  and  cl  by  constructing  (at  will)   a 
force  triangle  as  abO.    This  is  accomplished  by  assuming  the  intensity  of 
one  of  the  components,  say  c,  and  laying  off  aO  from  0  in  any  convenient 
direction  to  represent  this  intensity,  and  then  from  a  laying  off  ab  equal 
and  parallel  to  PI,  and  drawing  bO  to  complete  the  triangle,  we  have  cl, 
the  other  component,  fully  given  by  this  last  line  bO.     Then  from  0  pro- 
long the  line  of  action  of  the  component  cl  until  it  intersects  the  line  of 
action  of  the  force  P2  at  0' '.     Then  at  this  point  resolve  P2  into  two 
components  such  that  one  of  them  will  be  equal  and  opposite  to  the  com- 
ponent cl  at  0  and  act  along  the  same  line  00'.     Then  there  will  be  a 
component  force  cl  at  0  and  an  equal  and  opposite  component  force  cl 
at  0',  which  will  balance  each  other.     The  component  cl  at  0'  being 
designated,  we  obtain  the  other  component  of  P2  which  we  will  call  c2, 


36 


STRUCTURAL  ENGINEERING 


by  constructing  the  force  triangle  deO' ',  which  is  constructed  by  laying  off 
dO'  equal  and  parallel  to  cl  and  de  equal  and  parallel  to  P2,  and  drawing 
eO'  for  completion.  Then  c2  is  given  by  this  last  line  eO' '.  Now  from 
0'  prolong  the  line  of  action  of  the  component  force  c2  until  it  intersects 
the  line  of  action  of  the  force  P3  at  0".  Then  at  this  point  (O")  resolve 
P3  into  two  components  such  that  one  of  them  will  be  equal  and  opposite 


Fig.  32 


to  the  component  c2  at  0'  and  act  along  the  same  line  O'O".  Then  there 
will  be  a  component  force  c2  at  0'  and  an  equal  and  opposite  component 
force  c2  at  0"  which  will  balance  each  other.  The  one  component  force 
c2  at  0"  being  designated,  the  other  component  of  P3,  which  we  will  call 
c3,  we  obtain  by  constructing  another  force  triangle  ghO" ',  where  c3  is 
given  by  the  line  hO". 

Now  it  will  be  observed  that  the  three  forces  PI,  P2,  and  P3  at  (a) 


FUNDAMENTAL  ELEMENTS  37 

have  each  been  resolved  into  two  component  forces  which  replace  the 
original  force  in  each  case,,  and  that  these  component  forces  are  all  mutu- 
ally balanced,,  except  c  (at  0)  and  c3  (at  0"),  so  that  the  three  original 
forces,,  PI,  P2,  and  P3,  are  really  replaced  by  these  two  component  forces, 
c  and  c3.  So,  evidently,  if  we  determine  the  resultant  of  these  two  com- 
ponent forces  in  its  true  position,  we  will  have  the  resultant  of  the  three 
original  forces,  PI,  P2,  and  P3,  fully  determined  in  its  true  position,  as 
evidently  the  resultant  in  the  two  cases  must  be  identical. 

Now  as  the  two  component  forces  c  and  c3  are  intersecting  forces, 
their  resultant  is  readily  determined  in  its  true  position  by  prolonging 
their  lines  of  action  until  they  intersect  at  /  and  constructing  the  force 
triangle  Ihm,  where  their  resultant  is  fully  given  in  its  true  position  by 
the  line  Im.  This  resultant  is  the  same  in  every  respect  as  the  resultant 
of  the  original  forces  PI,  P2,  and  P3,  and  it  may  be  assumed  to  be  applied 
at  any  point  along  its  line  of  action  yy. 

Another  way  of  determining  the  resultant  is  as  follows: 

Instead  of  resolving  the  forces  PI,  P2,  and  P3  into  components,  as 
was  done  above,  we  can  select  any  point  on  the  line  of  action  of  one  of  the 
forces,  as  point  S  (really  the  same  as  we  did  before)  on  the  line  of  action 
of  PI  (at  (a))  and  apply  two  forces  /  and  /I  such  that  the  force  PI  will 
be  balanced  by  them.  This  is  accomplished  by  assuming  the  intensity 
and  direction  of  one  of  the  forces  and  then  constructing  a  force  triangle 
to  determine  the  intensity  and  direction  of  the  other.  Thus,  beginning  at 
S,  lay  off  Sn  (at  will)  to  represent  /  in  intensity  and  direction.  Then 
from  n  lay  off  no  equal  and  parallel  to  PI  and  draw  oS  for  completion  of 
the  triangle  and  we  have  /I  represented  in  intensity  and  direction  by  this 
last  line  oS.  As  the  three  forces  at  S  are  in  equilibrium,  they  will  act  in 
the  same  direction  around  the  triangle  Sno  as  indicated,  which  is  in 
accordance  with  Art.  42.  Next,  from  S  prolong  the  line  of  action  of  the 
force  /I  until  it  intersects  the  line  of  action  of  the  force  P2  at  S'.  Then 
at  this  point  apply  two  forces  such  that  they  will  balance  P2  and  one  of 
them  be  equal  and  opposite  to  the  force  /I  at  S  and  act  along  the  same 
line  SS'.  Then  there  will  be  a  force  /I  at  S  and  an  equal  and  opposite 
force  fl  at  S'  which  will  balance  each  other.  The  force  fl  at  S'  being 
known,  the  other  balancing  force  at  S',  which  we  will  designate  as  /2,  we 
determine  by  constructing  the  force  triangle  S'pr  in  the  same  manner  as 
was  explained  in  the  case  of  triangle  snO  at  S.  Then  from  S'  prolong 
the  line  of  action  of  the  force  f2  until  it  intersects  the  line  of  action  of 
P3  at  S".  Then  at  this  point  (£")  apply  two  forces  such  that  they  will 
balance  P3  and  one  of  them  be  equal  and  opposite  to  f2  at  S'  and  act 
along  the  same  line  S'S".  Then  there  will  be  a  force  f2  at  S'  and  an 
equal  and  opposite  force  f2  at  $"  which  will  balance  each  other.  The 
force  f2  at  S"  being  known,  the  other  balancing  force,  which  we  will 
designate  as  /3,  at  S",  we  determine,  by  constructing  another  force 
triangle  S"st,  where  /3  is  given  by  the  line  S"t. 

Now  it  will  be  observed  that  each  of  the  three  forces  PI,  P2,  and  P3 
is  balanced,  that  is,  held  in  equilibrium  by  two  forces  applied  in  each 
case  for  that  purpose,  and  such  being  the  case,  the  system  will  be  in 
equilibrium.  But  it  will  be  observed  that  all  of  the  balancing  forces  are 
mutually  balanced  except  the  force  /  at  S  and  /3  at  S".  Then,  evidently, 
these  two  forces,  /  and  /3,  hold  the  three  forces  PI,  P2}  and  P3  in 


38  STRUCTURAL  ENGINEERING 

equilibrium,  in  which  case  the  resultant  of  the  forces  /  and  /3  will  be 
an  equal  and  opposite  conspiring  force  to  the  resultant  of  the  forces 
PI,  P2,  and  P3  (see  Art.  42).  So  if  we  determine  the  resultant  of  the 
two  forces  /  and  /3  in  its  true  position,  we  will  have  a  force  the  same  in 
position  and  in  intensity  as  the  resultant  of  PI,  P2,  and  P3,  but  which 
acts  in  the  opposite  direction,  so  that  the  resultant  of  the  forces  PI,  P2, 
and  P3  will  thus  be  relatively  determined  in  its  true  position.  To  deter- 
mine the  resultant  of  the  forces  /  and  /3,  prolong  their  lines  of  action 
until  they  intersect  at  /'  and  construct  the  force  triangle  I'k'm'  and  we 
have  their  resultant  fully  represented  by  the  line  I'm'. 

The  only  difference  in  the  two  methods  shown  above  is  due  to  the  com- 
ponents being  reversed  in  direction  in  the  second  method  so  that  they  are 
really  balancing  forces  instead  of  actual  components.  Otherwise  the  two 
methods  are  identical.  Either  of  the  broken  lines  xOO'0"z  or  vSS'S"u 
would  be  known  as  an  Equilibrium  Polygon,  and  the  lines  xO,  00', 
O'O",  and  0"z,  or  vS,  SS',  S'S",  and  S"u  would  be  known  as  their 
respective  segments. 

It  is  evident  that  the  resultant  of  any  number  of  such  forces  as  PI. 
P2  and  P3  can  be  determined  in  its  true  position  in  the  same  manner  as 
shown  above  for  these  three  forces,  but  in  practice  the  same  thing  is 
accomplished  with  less  work  by  a  somewhat  different  method  of  pro- 
cedure, as  will  now  be  shown. 

Imagine  the  triangle  abO,  at  (a)  Fig.  32,  moved  bodily  and  parallel 
to  itself  to  the  position  abP  at  (b}.  Next  imagine  the  triangle  deO'  at 
(«)  moved  in  the  same  manner  to  the  position  beP,  d  coinciding  with  b 
and  0'  with  P,  and  likewise  imagine  the  triangle  ghO"  moved  in  the  same 
manner  to  the  position  ehP,  g  coinciding  with  e  and  0"  with  P.  Now  we 
have  the  three  force  triangles  abO,  deO' ,  and  ghO"  collected  into  a  single 
diagram  PabehP  (at  (&))  which  we  will  call  a  Ray  Diagram,  where  the 
line  abeh  is  the  load  line,  and  the  lines  aP,  bP,  eP,  and  hP  are  the  rays, 
and  the  point  P  is  the  pole.  It  will  be  observed  that  these  rays,  aP,  bP, 
eP,  and  hP  are  parallel,  respectively,  to  the  segments  xO,  00',  O'O", 
and  0"z  of  the  equilibrium  polygon  xOO'0"z,  at  (a).  So,  evidently,  if 
this  ray  diagram,  PabehP,  had  been  drawn  beforehand,  the  equilibrium 
polygon  xOO'0"z  could  have  been  constructed  by  beginning  at  0  and 
drawing  xO  parallel  to  the  ray  aP,  then  from  0  drawing  00'  parallel  to 
the  ray  bP,  and  likewise  from  0'  drawing  O'O"  parallel  to  the  ray  eP, 
and  then  from  0"  drawing  0"z  parallel  to  the  ray  hP.  Then  after 
having  drawn  the  equilibrium  polygon  xOO'0"z  in  this  manner,  the 
segments  xO  and  zO"  could  be  prolonged  until  they  intersect  at  /,  and 
thus  the  point  7  would  be  located,  which  is  one  point  upon  the  line  of 
action  of  the  desired  resultant.  But  the  things  lacking  would  be  the 
intensity,  direction,  and  direction  of  action  of  the  resultant.  But  it  will 
be  observed  at  (b)  that  by  drawing  the  line  ah  we  will  have  a  force 
polygon  abeha  which  represents  the  forces  PI,  P2,  and  P3,  and  their 
resultant,  each  in  intensity,  direction,  and  direction  of  action.  Then  the 
intensity,  direction,  and  direction  of  action  of  the  desired  resultant  is 
given  by  the  line  ah.  So  by  drawing  through  /  a  line  parallel  to  this  line 
ah  we  would  have  the  line  of  action  of  the  desired  resultant,  and  hence 
the  resultant  would  be  fully  determined,  as  its  intensity  and  direction  of 


FUNDAMENTAL  ELEMENTS  39 

action,  as  stated  above,  as  well  as  its  direction,  are  given  by  the  line  ah 
in  the  force  polygon  abeha  at  (6). 

Now,  according  to  Art.  39,  in  any  case  the  force  polygon  represent- 
ing any  two  or  more  forces  and  their  resultant  can  always  be  drawn.  So 
suppose  in  the  above  case  the  force  polygon  abeha  at  (6)  be  the  very 
first  thing  drawn.  Then  we  would  have  the  intensity,  direction,  and 
direction  of  action  of  the  resultant  of  the  three  forces  PI,  P2,  and  P3 
given  by  the  line  ah,  and  the  only  thing  lacking  would  then  be  the  loca- 
tion of  the  line  of  action  of  the  resultant,  which  can  be  obtained  as  above 
by  constructing  an  equilibrium  polygon.  But  we  would  have  so  far  only 
the  force  polygon  abeha  constructed.  Now  if  the  point  P  at  (&)  and  the 
point  0  at  (a)  were  given,  we  could  draw  the  rays  aP,  bP,  etc.,  and 
beginning  at  0  we  could  draw  the  equilibrium  polygon  xOO'0"z,  as 
explained  above,  and  thus  obtain  the  desired  line  of  action  yy.  However, 
it  is  not  necessary  to  have  the  location  of  either  of  these  particular  points 
given,  as  the  location  of  0  was  arbitrarily  assumed  in  the  first  place,  and 
the  position  taken  by  the  equilibrium  polygon  xOO'0"z  was  governed  by 
the  arbitrarily  constructed  triangle  abO.  So,  evidently,  we  might  just  as 
well  assume  the  pole  P  and  draw  the  rays  of  a  ray  diagram  and  construct 
an  equilibrium  polygon  accordingly  by  starting  at  any  convenient  point 
on  the  line  of  action  of  one  of  the  forces. 

As  an  example,  let  AB,  Fig.  33,  represent  a  body  acted  upon  by  four 
forces,  PI,  P2,  P3,  and  P4.  First  construct  the  force  polygon  abcdea,  at 
(fr),by  drawingafc  equal  and  parallel 
to  PI,  and  be  equal  and  parallel  to 
P2,  and  so  on  as  explained  in  Art.  39, 
and  we  have  the  resultant  of  the  four 
forces  PI  ...  P4  given  in  intensity, 
direction,  and  direction  of  action  by 
the  line  ae.  Then  to  obtain  its  line 
of  action,  take  any  convenient  point 
P  as  a  pole  (at  (&))  and  draw  the 
rays  aP,  bP,  etc.,  thereby  obtaining 
the  ray  diagram  PabcdeP.  Then 
construct  a  corresponding  equilibrium 
polygon  at  (a)  as  xOO'0"0'  "z, 
which  is  accomplished  in  the  follow- 
ing manner:  First  select  any  con- 
venient point  on  the  line  of  action  of 
one  of  the  forces,  say,  point  0  on  the  Fig.  33 

line  of  action  of  PI.    Then,  from  this 

point  draw  the  segments  xO  and  00'  parallel  to  the  rays  aP  and  bP, 
respectively.  Then  from  0'  draw  the  segment  O'O"  parallel  to  the  ray 
cP.  Then  from  0"  draw  the  segment  0"0'  "  parallel  to  the  ray  dP. 
Then  from  0' "  draw  the  segment  zOf "  parallel  to  the  ray  eP.  Then 
prolong  the  segments  xO  and  zO' "  until  they  intersect  at  I,  and  through 
this  point  I  draw  the  line  yy  parallel  to  the  line  ae  in  the  force  polygon 
abcdea,  and  it  will  be  the  desired  line  of  action  of  the  resultant  of  the 
four  forces,  and  hence  the  resultant  is  fully  determined. 

If  the  forces  be  absolutely  parallel,  the  load  line  will  be  a  straight 


40 


STRUCTURAL  ENGINEERING 


line,,  and  the  line  representing  their  resultant  will  coincide  with  this  line, 
but,  however,  the  work  of  determining  the  resultant  is  the  same  as  shown 
above.  For  example,  let  AB,  at  (a)  Fig.  34,  represent  a  body  acted  upon 
by  four  parallel  forces  PI  ...  P4.  We  construct  the  load  line  abode  at 
(b)  by  laying  off  the  forces  in  consecutive  order  as  shown.  Then  assume 
the  pole  P  and  construct  the  ray  diagram  PabcdeP  and  draw  the 
equilibrium  polygon  xOO'O"0'  "  2  at  (a)  as  before  and  locate  the  point 
I,  through  which  draw  the  line  of  action  yy  of  the  resultant  parallel  to  the 
load  line  ae. 


1  \"   i 

K 

Al 

^ 

ID 

f\             1s 

\                   1      x 

/ 

/'  ^o*  * 

>vx    Ns 

'y  (a) 


Fig.   34 

The  method  of  procedure  just  shown  is  quite  satisfactory  in  practice, 
but  the  student  should  not  acquire  the  habit  of  constructing  the  force 
polygon  and  the  ray  diagram  and  then  the  corresponding  equilibrium 
polygon  without  fully  recognizing  the  exact  significance  of  each  step. 

In  Fig.  32  the  force  triangles  were  constructed  on  the  segments  of 
the  equilibrium  polygons,  while  in  Figs.  33  and  34  the  force  triangles 
were  constructed  on  the  load  lines  of  the  force  polygons,  but  the  principle 
involved  is  just  the  same  in  either  case.  In  the  case  shown  in  Figs.  33 
and  34  we  really  resolved  each  of  the  forces  PI,  P2,  P3,  and  P4  into  two 
components  by  constructing,  respectively,  the  triangles  abP,  bcP,  cdP, 
and  deP.  However,  it  is  easy  to  overlook  this  fact  when  the  triangles 
are  constructed  by  merely  drawing  the  rays  aP,  bP,  etc.  If  we  consider 
the  forces  resolved  into  components,  as  is  the  usual  custom,  the  com- 
ponent in  each  case  will  act  around  the  force  triangle  in  the  opposite 
direction  to  that  of  the  force,  while  if  we  reverse  the  direction  of  the 
components,  they  become  balancing  forces,  in  which  case  they  will  act  in 
the  same  direction  around  the  triangle  as  the  force  (see  Arts.  39  and  42). 
Take,  for  example,  the  case  shown  in  Fig.  33 ;  the  two  sides  aP  and  bP  of 
the  force  triangle  abP,  at  (6),  give  directly  either  the  intensity  and 
direction  of  two  components  of  PI  or  of  two  balancing  forces.  If  we 
assume  components,  the  one  represented  by  the  side  bP  will  act  from  P 
to  b,  but  if  we  assume  balancing  forces,  the  one  represented  by  the  side 
aP  will  act  from  P  to  a,  while  the  one  represented  by  the  side  bP  will  act 
from  b  to  P.  The  same  is  true  of  all  the  other  force  triangles  bcP,  cdP, 
and  deP  shown  at  (b).  In  constructing  an  equilibrium  polygon  it  really 
makes  no  difference  whether  we  consider  the  forces  resolved  into  com- 
ponents or  balanced  by  forces  applied  for  that  purpose,  as  the  final  result 
will  be  just  the  same. 


FUNDAMENTAL  ELEMENTS  41 

In  constructing  an  equilibrium  polygon  after  the  ray  diagram  is 
completed,  we  virtually  transfer  the  two  components  or  the  two  balancing 
forces,  as  the  case  may  be,  of  each  force  to  a  point  upon  its  actual  line  of 
action.  In  order  to  show  what  is  really  involved  in  the  construction,  take 
for  an  example  the  case  shown  in  Fig.  33.  As  the  forces  here  were 
assumed  to  be  resolved  into  components,  the  components  will  in  each  case 
act  around  the  force  triangle  in  the  opposite  direction  to  that  of  the  force. 
Now  beginning  with  PI,  we  can  assume  its  own  components,  represented 
by  the  sides  aP  and  bP  of  the  force  triangle  abP  at  (6),  as  applied  at  any 
point  0  upon  the  line  of  action  of  PI.  Then  as  the  component  represented 
by  the  side  aP  acts  from  a  to  P,  a  line  drawn  from  0  in  this  direction  and 
parallel  to  aP  will  represent  that  component  as  being  applied  at  0,  and 
we  thus  obtain  the  segment  xO  of  the  equilibrium  polygon.  The  other 
component  of  PI,  which  is  represented  by  the  side  bP,  acts  from  P  to  b. 
Then  a  line  drawn  from  0  in  this  direction  and  parallel  to  bP  will  repre- 
sent that  component  as  being  applied  at  0,  and  we  thus  obtain  the  seg- 
ment 00' ,  which  is  really  the  line  of  action  of  this  other  component  of  PI 
drawn  only  to  0',  where  it  intersects  the  force  P2.  We  can  assume  the 
two  components  of  P2,  represented  by  the  sides  bP  and  cP  of  the  force 
triangle  bcP  at  (6),  applied  at  any  point  upon  the  line  of  action  of  P2  the 
same  as  in  the  case  of  the  two  components  of  PI.  Then,  evidently,  we 
can  assume  them  applied  at  0'  as  well  as  at  any  other  point.  Now  the 
component  of  P2  represented  by  the  side  bP  of  the  force  triangle  bcP 
at  (6)  acts  from  b  to  P.  Then  a  line  drawn  from  0'  in  this  direction  and 
parallel  to  bP  will  represent  that  component  of  P2  as  being  applied  at 
Of ',  and  we  thus  obtain  the  segment  00'  again.  The  other  component  of 
P2  represented  by  the  side  cP  acts  from  P  to  c.  Then  a  line  drawn  in 
this  direction  from  0'  and  parallel  to  cP  will  represent  that  component 
applied  at  0',  and  we  thus  obtain  the  segment  O'O" ,  which  is  really  the 
line  of  action  of  this  other  component  of  P2  drawn  only  to  0"  where  it 
intersects  the  force  P3.  The  two  components  of  P3,  represented  by  the 
sides  cP  and  dP  of  the  force  triangle  cdP  at  (6),  can  be  assumed  to  be 
applied  at  0" ',  and  then  the  two  components  of  P4,  represented  by  the 
sides  dP  and  eP  of  the  force  triangle  deP,  can  be  assumed  to  be  applied 
at  0'  ",  and  the  discussion  of  the  previous  cases  will  apply  to  each  of 
these  cases. 

The  direction  of  action  of  the  components  of  each  force  can  be 
indicated  on  the  rays  of  the  ray  polygon,  and  on  the  segments  of  the 
equilibrium  polygon  as  in  Fig.  33,  but  usually  we  omit  such  indications, 
because  such  are  usually  not  necessary  to  any  great  extent — however,  that 
is  a  minor  item. 

After  an  equilibrium  polygon  is  drawn  for  a  system  of  forces  as 
shown  in  Fig.  33,  the  resultant  of  any  number  of  these  forces,  if  taken  in 
consecutive  order,  can  be  determined  as  well  as  the  resultant  of  all  of 
them.  For  example,  take  the  three  consecutive  forces  PI,  P2,  and  P3 
(Fig.  33).  By  drawing  the  line  ad  (at  (b)),  we  have  the  force  polygon 
dbcda,  wherein  the  resultant  of  the  three  forces  is  given  in  intensity, 
direction,  and  direction  of  action  by  this  line  ad.  By  prolonging  the 
segment  0"0'  "  until  it  intersects  the  segment  xO  at  /',  we  have  one 
point  on  the  line  of  action  of  the  resultant  of  these  three  forces,  PI,  P2, 
and  P3.  Then  by  drawing  a  line  through  /'  parallel  to  the  line  ad,  we 


42 


STRUCTURAL  ENGINEERING 


obtain  the  desired  line  of  action  of  their  resultant,  and  thus  we  have  the 
resultant  of  the  three  forces  determined.  As  another  case,  the  resultant 
of  the  two  forces  P2  and  P3  can  be  determined  by  drawing  a  line  bd 
(at  (fc))  p^d  prolonging  the  segments  0"0'  "  and  00'  until  they  intersect 
at  I". 

Second  Method:  The  following  graphical  method,  which  may  be 
designated  as  the  Proportional  Triangle  Method,  can  be  used  to  advantage, 
quite  often,  in  determining  the  resultant  of  parallel  forces : 

Let  PI  and  P2  (Fig.  35)  represent  any  two  parallel  forces,  and  let 


Fig.  35 


R  represent  their  resultant.  Let  a  line  ab  be  drawn  perpendicular  to  the 
two  forces  and  their  resultant,  intersecting  the  resultant  at  o.  As  there  is 
no  question  as  to  the  intensity  of  the  resultant  being  equal  to  the  sum  of 
the  forces,  that  is,  ^tJ  =  Pl  +  P2,  and  the  direction  being  the  same  as  the 
forces,  it  remains  only  to  determine  the  point  o  through  which  the  line  of 
action  of  the  resultant  passes.  This  can  be  obtained  by  the  following 
construction:  From  b  draw  a  line  bx,  making  any  convenient  angle  with 
the  line  ab.  Then  from  b  lay  off,  to  any  convenient  scale,  bc  =  Pl;  and 
from  c  lay  off,  to  the  same  scale,  cd=P2.  Then  draw  ad  and  through  c 
draw  a  line  parallel  to  ad,  and  where  it  intersects  the  line  ab  will  be  the 
required  point  o  through  which  the  resultant  passes.  The  same  can  be 
accomplished  by  drawing  the  line  ay  from  a  and  laying  off  the  loads  as 
indicated  and  then  drawing  the  lines  fb  and  eo. 

The  above  is  readily  seen  to  be  true,  for  taking  moments  about  b, 
according  to  Art.  41,  we  have 

obxR  =  Plxab  or  ob  =  — — — 

which  is  the  distance  of  the  point  o  from  b;  but  from  similar  triangles, 
bco  and  bdat  we  have 

ob         cb  PI 


or 


Plxab 


and  thus  we  have  an  identity. 


FUNDAMENTAL  ELEMENTS  43 

In  case  there  are  more  than  two  forces,  the  resultant  of  any  two  is 
determined  as  above  and  this  resultant  is  then  considered  with  one  of  the 
remaining  forces,  and  so  on.  For  example,  let  PI,  P2,  and  P3  (Fig.  36) 


represent  three  parallel  forces.  First  draw  a  line  as  abc  perpendicular  to 
the  forces.  Then  from  one  of  the  points  a,  b,  or  c,  say  a,  draw  a  line  axt 
making  any  convenient  angle  with  the  line  abc.  Then  lay  off  ae  =  P2  and 
eg  =  Pl.  Then  draw  gb  and  through  e  draw  a  line  parallel  to  gb,  and  the 
point  k  where  this  line  intersects  the  line  abc  is  a  point  on  the  line  of 
action  of  the  resultant  of  the  two  forces  PI  and  P2.  Then  from  this 
point  draw  a  line  ky,  making  any  convenient  angle  with  the  line  abc. 
Then  lay  off  kh  =  P3  and  hm  =  Pl+P2.  Then  draw  cm  and  through  h 
draw  a  line  parallel  to  cm,  and  the  point  o  where  this  line  intersects  the 
line  abc  is  a  point  on  the  line  of  action  of  the  resultant  of  the  three 
parallel  forces. 

It  is  important  to  note  that  the  force  from  which  the  diagonal  line 
is  drawn  is  laid  off  last  on  that  line.  Thus  in  Fig.  36,  eg  =  PI  and 
hm  =  PI  +  P2. 

In  cases  where  the  forces  are  in  groups,  as  locomotive  wheel  loads, 
the  method  is  a  very  convenient  one,  as  the  resultant  of  each  group  is 
known  from  observation  and  thus  the  forces  really  considered  in  the  con- 
struction are  quickly  reduced  to  only  a  few. 

46.  Analytical  Determination  of  the  Resultant  of  Parallel 
Forces.— It  is  evident  that  the  resultant  of  two  or  more  parallel  forces 
must  act  in  the  same  direction  as  the  forces,  and  with  an  intensity  equal 
to  their  algebraic  sum;  otherwise  it  would  not  produce  the  same  effect  as 
the  forces  themselves.  So  the  intensity,  direction,  and  direction  of  action 
of  the  resultant  of  two  or  more  parallel  forces  are  really  known  directly 
from  the  forces  themselves,  and  such  being  the  case,  the  problem  involved 
in  determining  their  resultant  really  reduces  to  the  locating  of  the  line  of 
action  of  the  resultant. 

Let  PI  .  .  .  P5  represent  five  parallel  forces  acting  upon  the  body 
AB  (Fig.  37),  and  let  R  represent  their  resultant.  Then  we  have 
#  =  P1+P2  +  P3+P4  +  P5.  According  to  Art.  41,  the  moment  of  this 
resultant  about  any  point  is  equal  to  the  algebraic  sum  of  the  moments  of 
the  five  forces  about  the  same  points.  Then  taking  moments  about  some 
point  0,  we  have 

aPl  +  bP2  +  cP3  +  dP4:  +  eP5  =  R*, 
from  which  we  get 

aPl  +  bP2  +  cP3  +  dP4  4-  eP5 


R  P1+P2  +  P3  +  P4  +  P5 


44 


STRUCTURAL  ENGINEERING 


So  then  we  have  the  line  of  action  of  R  located  in  reference  to  O. 

In  practice  it  is  usual  to  take  moments   about  one  of  the   forces, 


At 


Fig.  37 


thereby  shortening  the  work  by  eliminating  the  moment  of  that  force. 
Thus  taking  moments  about  PI  (Fig.  38),  we  have 


_ 
" 


In  case  the  moments  about  one  point  are  known,  the  moments  about  any 
other  point  can  be  determined  by  either  increasing  or  diminishing  the 
known  moments  by  the  sum  of  the  forces  multiplied  by  the  common  dif- 
ference of  lever  arms.  For  example,  the  moments  about  the  point  z 
(Fig.  38)  are  equal  to  the  moments  about  PI  plus  the  sum  of  the  forces 


d 

0 

oc 

p  b  . 

._*. 

* 

p— 

I 

* 

\ 

\ 

* 

^  1  n 

Fig.  38 


Fig.   39 


PI  .  .  .  P5  multiplied  by  k.  That  is,  the  moments  about  z  are  equal  to 
6P2-l-cP3  +  dP4  +  eP5+(Pl+P2  +  P3  +  P4  +  P5)Ar.  This  is  readily  seen 
to  be  true,  for  let  M  be  the  moment  of  a  force  P  about  A  (Fig.  39)  and  let 
M'  be  its  moment  about  B;  then  we  have  M  —  aP  and  M'  =  P(a  +  b)  = 
aP  +  bP  =  M  +  bP.  It  is  readily  seen  that  this  is  as  true  for  a  number 
of  forces  as  it  is  for  one.  The  product  of  the  sum  of  the  forces  and 
common  difference  of  lever  arms  would  be  subtracted  from  the  moments 
in  case  the  lever  arms  became  shortened  by  the  transferring  of  the  center 
of  moments. 

47.  Center  of  Gravity. — The  most  common  case  of  parallel  forces 
is  that  of  gravity,  which  acts  with  equal  intensity  upon  each  ultimate 
part  of  material  contained  in  a  body,  regardless  of  the  kind  of  material. 
In  case  of  an  individual  body,  the  line  of  action  of  the  resultant  of  the 
gravity  forces  passes  through  what  is  known  as  the  center  of  gravity  of 


FUNDAMENTAL  ELEMENTS  45 

the  body,  but  when  more  than  one  body  is  considered,  the  center  of  gravity 
of  the  bodies  considered  as  a  group  is  the  same  as  the  resultant  of  so  many 
parallel  forces,  the  weight  of  each  body  being  considered  as  a  force  acting 
through  the  center  of  gravity  of  the  respective  body. 

We  can  consider  symmetrical  bodies  as  being  made  up  of  parallel 
layers  of  homogeneous  material  wherein  the  centers  of  gravity  of  the 
layers  coincide  each  with  each,  and  such  being  the  case,  we  need  treat 
only  one  layer,  which  we  may  consider  as  a  plane  without  thickness,  from 
which  results  the  common  practice  of  treating  area  instead  of  volume, 
weight,  or  mass. 

In  case  the  material  composing  a  body  be  symmetrical  about  a  plane, 
there  is  no  question  but  that  the  center  of  gravity  of  the  body  will  be  in 
that  plane;  and  hence,  if  the  material  be  symmetrical  about  three  or  more 
planes  intersecting  in  a  point,  their  point  of  intersection  will  undoubtedly 
be  the  center  of  gravity  of  the  body.  So  it  is  evident  that  the  centers  of 
gravity  of  many  bodies  and  plane  figures  are  known  from  mere  inspection 
of  their  state  of  symmetry.  Thus,  the  center  of  gravity  of  a  sphere  is  at 
its  center,  a  line  at  its  middle  point,  and  we  know  the  center  of  gravity 
of  a  circle,  cylinder,  cube,  an  ellipse,  etc.,  from  the  same  deduction.  In 
fact,  whatever  the  method  employed  in  determining  the  centers  of  gravity 
of  bodies  and  plane  figures,  we  assume  the  center  of  gravity  of  certain 
elements  as  being  predetermined  by  mere  symmetry. 

The  determination  of  the  center  of  gravity  of  loads  in  groups  and  the 
center  of  gravity  of  the  cross-section  of  individual  members  of  structures 
are  the  principal  center  of  gravity  problems  occurring  in  structural 
engineering. 

We  can  determine  the  center  of  gravity  of  any  group  of  loads  in  the 
same  manner  as  the  resultants  of  parallel  forces  were  determined  in  Arts. 


o 


Fig.  40  Fig.   41 

45  and  46.  The  analytical  method  outlined  in  Art.  46  is  most  used  as  it  is 
usually  more  convenient  than  any  other  method.  As  an  example,  let 
PI  .  .  .  P4  (Fig.  40)  represent  four  loads.  Taking  moments  about  one 
of  the  end  loads,  either  PI  or  P4,  say  P4,  we  have 

cPl+bP2+aP3 
"P1  +  P2  +  P3  +  P4  ' 

where  x  is  the  distance  the  center  of  gravity  of  the  four  loads  is  from  P4. 
In  case  of  the  cross-section  of  a  member,  we  deal  with  the  area 
instead  of  the  weight.  For  example,  let  the  sketch  in  Fig.  41  represent 
the  cross-section  of  a  plate  and  an  angle  which  is  riveted  to  the  plate. 
Let  A  be  the  area  of  the  plate,  and  let  A'  be  the  area  of  the  angle,  and  let 
a  and  b  be  the  distance  from  the  bottom  edge  of  the  plate  to  the  center 
of  gravity  of  the  plate  and  angle,  respectively.  Then  taking  moments 
about  the  bottom  edge  of  the  plate,  we  have 


46 


STRUCTURAL  ENGINEERING 


or 


A'b  )  A  a    — 
—  —  -  ~r=x> 
A    -{-A 


where  x  is  the  distance  from  the  bottom  edge  of  the  plate  to  the  center  of 
gravity  of  the  total  cross-section  of  the  plate  and  angle  combined.  The 
work  could  be  shortened  by  taking  moments  about  the  center  of  gravity 
of  the  plate  or  angle,  thus  following  out  the  same  scheme  as  was  used  in 
the  case  of  loads  (Fig.  40). 

The  center  of  gravity  of  plain  areas  can  be  determined  quite  readily 
by  the  aid  of  calculus,  wherein  the  general  formula  is 


For  example,  let  Fig.  42  represent  a  rectangle  having  a  height  h  and 
width  6.     Taking  moments  about  the  lower  edge  of  the  rectangle,  we  have 

*hbydy  =  bh2  _  bh2      h 
~~A 2A  ~2bh~2' 

That  is,  the  center  of  gravity  of  the  rectangle  from  its  lower  edge  is  equal 
to  one-half  of  its  height,  which  we  really  knew  from  mere  inspection. 


Fig.   42 


Fig.   43 


As    another    example,    let    Fig.    43    represent   a    triangle.      Taking 
moments  about  a  line  AB  parallel  to  its  base,  we  have 


~=  [yda  _ 


2bhs     2 


h  A 

which  is  the  distance  of  the  center  of  gravity  of  the  triangle  from  the 
axis  AB. 

Problem  7.  Determine  the  center  of  gravity  in  reference  to  the 
horizontal  axis  of  the  section  shown  in  Fig.  44,  which  is  composed  of 
3—14"  x  \"  plates  and  2—6"  x  6"  x  \"  angles. 

Solution:  Taking  moments  of  the  areas  about  the  center  of  gravity 
of  the  top  plate,  we  have 

32.5  *  =  0.5  x  7  +  1  x  7  +  2.93  x  11.5, 
-    44.19 


or 


x  = 


32.5 


=  1.36". 


Hence  the  center  of  gravity  of  the  section  is  1.36"  below  the  center  of  the 
top  plate,  or  0.11"  below  the  back  of  the  angles.     The  numerical  quanti- 


FUNDAMENTAL  ELEMENTS 


47 


ties  in  this  and  the  preceding  problem  can  be  verified  by  referring  to  the 
tables  in  the  back  of  this  book. 

Problem  8.  Determine  the  center  of  gravity  in  reference  to  the 
horizontal  axis  of  the  section  shown  in  Fig.  45,  which  is  composed  of 
2 — 15"  x  33*  channels  and  one  cover  plate  22"  x  J". 


3- 


I*  ^ 

1        K* 


33*-" 


Fig.   44 


V 

Fig.   45 


Solution:     Taking  moments  of  the  areas  about  the  center  of  gravity 
of  the  cover  plate,  we  have 

30.8^=7.75x19.8, 
-    7.75x19.8 


or 


x  — 


30.8 


=  4.98". 


That  is,  the  center  of  gravity  of  the  entire  section  is  4.98"  below  the 
center  of  the  cover  plate,  or  4.73"  from  the  back  of  the  top  flanges  of  the 
channels.  The  center  of  gravity  here  lies  on  the  horizontal  line  marked 
gg,  which  is  known  as  the  gravity  line  or  gravity  axis. 

It  is  readily  seen  that  the  center  of  gravity  of  the  above  section  in 
reference  to  the  vertical  axis  would  lie  on  the  vertical  line  vv  passing 
through  the  center  of  the  cover  plate;  however,  the  center  of  gravity  of 
any  section  in  reference  to  a  vertical  axis  can  be  obtained  by  taking 
moments  about  some  vertical  line  in  the  same  manner  as  shown  above  for 
the  horizontal  axis. 

48.  Inertia. — It  is  an  observed  fact  that  it  always  requires  a  force 
to  move  a  body  when  at  rest,  to  stop  it  when  in  motion,  or  to  change  its 
motion  in  any  respect.  This  would  be  true  even  if  friction  and  all  other 
external  resistances  were  removed.  Hence,  it  is  evident  that  a  body  in 
itself  offers  resistance  to  every  change  of  motion;  otherwise,  it  would 
require  no  force  to  change  the  motion  of  the  body  if  all  the  external 
resistances  were  absent.  This  property  which  bodies  have  in  themselves 
of  offering  resistance  to  change  of  motion  is  known  as  Inertia.  As  an 
example,  a  body  lying  upon  an  absolutely  smooth  horizontal  plane  would 
offer  resistance  to  any  change  of  motion  along  the  plane  simply  by  virtue 
of  its  inertia.  The  actual  force  exerted  would  be  directly  proportional 
to  the  mass  of  the  body  concerned,  and  to  the  change  of  motion,  or  in 
other  words  to  the  acceleration  produced.  Let  W  be  the  weight  of  the 
body  concerned,  and  let  a  be  the  acceleration  produced  in  feet  per  second. 


48 


STRUCTURAL  ENGINEERING 


Then  for  the  intensity  of  the  force   (in  pounds)   exerted,  we  have  the 
proportion 

F:  W::  a:  g, 
from  which  we  obtain 


8 


which  is  Formula  (A)  given  in  Art.  23. 

49.     Moment  of  Inertia  and  Radius  of  Gyration.  —  Let  AB  (Fig. 
46)    represent   a  very   thin   rectangular  plate  of   homogeneous   material. 

Suppose  the  plate  starts  to  rotating  about 
a  vertical  axis  YY.  The  resistance  offered 
by  all  such  strips  as  cd  (which  will  be 
due  to  their  inertia)  will  be  directly  pro- 
portional to  their  distance  out  from  the 
axis  YY,  as  their  relative  increment  of 
velocity  or  acceleration  will  be  directly 
proportional  to  that  distance.  Then  the 
intensity  of  the  resistance  of  any  strip 
will  be  directly  proportional  to  the  dis- 
tance of  the  strip  from  the  axis  and  to  the 
mass  of  the  strip.  Suppose  the  plate  made 
up  of  infinitesimal  strips  as  cd,  and  let  m  be  the  mass  of  each  strip.  Then 
the  force  resisting  the  motion  of  the  strip  out  unit  distance  from  the  axis 
will  be  (m)  1,  while  the  force  out  x  distance  will  be  mx,  and  the  moment  of 
it  about  the  axis  would  be  (mx)  (x)=mx2.  Then  evidently  the  moments 
about  the  axis  YY  of  all  the  resisting  forces  of  these  strips  will  be  ^mx2 
where  m  is  the  mass  of  each  strip,  which  was  assumed  to  be  the  same  for 
each  and  every  one,  and  x  the  distance  out  to  each  strip,  no  two  x's  being 
the  same.  The  expression  2moj2  is  known  as  the  Moment  of  Inertia,  which 
is  usually  indicated  by  I,  and  we  have  in  general 


Fig.   46 


In  the  case  of  the  above  plate  it  is  evident  that  all  of  its  mass  could 
be  concentrated  at  some  point  out  from  the  axis  YY  so  that  its  moment 
of  inertia  about  the  axis  would  be  the  same  as  that  of  the  plate.  This 
distance  out  would  be  known  as  the  Radius  of  Gyration  of  the  plate  in 
reference  to  the  axis  YY.  Then  we  have  the  general  equation 

7  =  Mr2, 

where  M  is  the  total  mass  of  the  plate  and  r  the  radius  of  gyration. 

The  mass  of  any  one  of  the  infinitesimal  strips  of  the  above  plate  is 
directly  proportional  to  da,  the  area  of  the  strip;  then  if  we  imagine  the 
plate  to  diminish  in  thickness  until  it  becomes  a  plane  without  thickness, 
we  have  for  its  moment  of  inertia 


from  which  we  obtain 


FUNDAMENTAL  ELEMENTS 


Let  a  be  the  length  of  the  plate  and  b  its  width ;  then  considering  the 
plate  as  a  plane,  its  moment  of  inertia  about  the  axis  YY  is 


1= 


Knowing  its   moment   of  inertia,   we   can   then   determine   its    radius   of 
gyration  from  the  above   formula,  I  =  Ar2.     Thus  substituting,  we  have 

•—-  =  bar2  or  r  — 
o 

The  determination  of  the  moment  of  inertia  and  radius  of  gyration 
of  the  cross-section  of  individual  members  of  structures  are  the  most 
common  problems  under  this  head  met  with  in  structural  engineering.  It 
is  really  the  case  of  determining  the  moment  of  inertia  and  the  radius  of 
gyration  of  a  plane,  or  as  we  may  put  it,  the  material  cut  by  a  plane. 
There  are  a  few  cases,  however,  where  it  is  necessary  to  consider  the 
mass,  but  in  such  cases  the  student  will  have  no  trouble  if  the  general 
formula,  I  —  Mr2,  given  above,  is  applied. 

$ 


Fig.  47 


Fig.  48 


Fig.   49 


PROBLEMS 


(1)      Moment  of  inertia  of  a  rectangle  in  reference  to  an  axis  GG 
(Fig.  47)  through  its  center  of  gravity. 
Solution:     Here  we  have 


-II 


'24   ^24 


bh? 
12 


(2)  Moment  of  inertia  of  a  triangle  about  its  base.  (Fig.  48.) 
Here  we  have 

b  fh       „  _bh* 

hjo  12 

In  case  the  axis  is  perpendicular  to  the  plane,  we  have  what  is  known 
as  the  Polar  Moment  of  Inertia.  For  example,  for  the  polar  moment  of 
inertia  of  a  circle  about  its  center  we  have  from  Fig.  49 


rR  CR 

2=   I     dax2=27r    I     x 

Jo  *Jo 


A  more  general  case  of  the  polar  moment  of  inertia  would  be  that  of 
a  rectangle.  Here  let  0  (Fig.  50)  be  the  axis  and  draw  XX  and  YY 
through  0  at  right  angles  to  each  other.  Then  we  have 


50 


STRUCTURAL  ENGINEERING 


which  is  the  moment  of  inertia  of  the  rectangle  about  the  axis  XX  plus 
the  moment  of  inertia  about  the  axis  YY.  This  gives  the  general  formula 
which  will  apply  to  any  plane  figure,  and  further  discussion  of  the  polar 
moment  of  inertia  is  unnecessary.  Its  principal  use  is  in  the  designing  of 
shafting.  The  polar  moment  of  inertia  is  usually  designated  by  the  letter 
J  instead  of  /,  which  is  known  as  the  Rectangular  Moment  of  Inertia. 
50.  Proposition. —  The  rectangular  moment  of  inertia  of  any  plane 
figure  about  any  axis  in  the  same  plane  as  the  figure  is  equal  to  its  moment 
of  inertia  about  a  parallel  axis  through  its  center  of  gravity ,  plus  its  area 
multiplied  by  the  square  of  the  distance  between  the  two  axes.  As  proof 
of  this,  let  ABCD  (Fig.  51)  represent  a  rectangle  whose  area  is  A=bh 


-9 

e 

r 

£& 

k 

X 

O 

/•x.  1 

x 

-n 

Y 

Fig.  50 

A 

3 

d 

a 

& 

- 

-dx 

oc 

L  h 

e 

Q 

0 

fe 

| 

9 

F 

ig.   51 

and  whose  moment  of  inertia  about  its  gravity  axis  gg  is  I' ' .  Then, 
according  to  the  above,  the  moment  of  inertia  of  the  rectangle  about  any 
axis  as  YY  is  /  =  I'  +  d2A.  This  is  shown  to  be  true  in  the  following 
manner:  The  moment  of  inertia  of  the  rectangle  about  the  axis  YY  is 

fo    •--,').  ..(1). 


/a 
1 


o 


Now  from  Fig.  51  we  have  a  =  d+  (h/2)  and  e  =  d-  (h/2).     Substituting 
these  values  of  a  and  e  in  (1)  we  have 


But  bh*/12  equals  the  moment  of  inertia  of  the  rectangle  about  its  gravity 
axis  gg.    Then  substituting  /'  for  bh3/12  and  A  for  bh,  we  have 


which  proves  the  above  proposition  in  the  case  of  a  rectangle. 

The  above  proposition  is  true  of  any  plane  figure.  This  can  be  shown 
by  following  out  the  same  scheme  as  above.  In  case  of  bodies,  the  only 
thing  different  is  that  we  would  consider  mass  instead  of  area  and  write 
the  formula 


instead  of 


=  I'  +  Ad 


CHAPTER  IV 

THEORETICAL  TREATMENT  OF  BEAMS 

51.  Classification. — Beams  are,  as  a  rule,  classified  according  co 
the  number  and  kind  of  supports  they  have.  Whenever  a  beam  simply 
rests  upon  a  support,  the  support  is  known  as  a  simple  support,  but  when 
the  beam  is  held  rigidly  by  a  support,  the  support  is  known  as  a  fixed 
support. 

Figures  52  to  55  show  the  most  common  types  of  beams.  The  canti- 
lever beam  shown  in  Fig.  52  has  one  fixed  support,  while  the  fixed  beam 


Fig.  52 


Fig.   53 


Fig.  54 


Fig.   55 


shown  in  Fig.  53  has  two  fixed  supports.  The  simple  beam  shown  in 
Fig.  54,  which  is  the  most  common  type,  has  two  simple  supports.  The 
continuous  beam  shown  in  Fig.  55  has  four  simple  supports;  however,,  any 
beam  with  more  than  two  supports  of  any  kind  is  a  continuous  beam. 

There  are  various  other  types  of  beams,  such  as  overhanging,  inclined, 
etc.,  but  they  are  all  really  modifications  of  the  above  types  and  will  be 
readily  recognized  as  such  without  any  preliminary  description. 

Beams  are  usually  supported  in  a  horizontal  position  and  support 
vertical  loads  which  produce  vertical  reactions.  This  will  be  understood 
to  be  the  case  unless  otherwise  stated. 

52.  Shearing  Stress  on  Beams. — Let  AB  (Fig.  56)  represent  an 
ordinary  rectangular  wooden  cantilever  beam  supported  in  a  horizontal 
position  by  having  the  end  B  built  into  a  wall.  Let  P  represent  a  load  at 
A  which  the  beam  supports  in  addition  to  its  own  weight.  Let  ab  and  cd 
represent  the  traces  of  two  imaginary  planes  passing  perpendicularly  to 
the  longitudinal  axis  of  the  beam,  mentally  separating  the  very  short 
rectangular  block  abed  from  the  other  parts  of  the  beam,  so  that  we  can 
think  of  this  block  as  being  a  separate  body. 

By  imagining  the  beam  to  be  cut  off  instantly  along  the  section  ab  we 
readily  realize  that  the  part  abA  of  the  beam  has  a  tendency  to  move 
down  vertically  and  the  load  P  with  it.  As  the  material  in  the  block  abed 
prevents  the  motion,  evidently  the  part  abA  of  the  beam  exerts  a  down- 
ward force  along  the  section  a b  upon  the  block  equal  to  the  combined 
weight  of  the  part  abA  of  the  beam  and  the  load  P.  Let  S  be  this  force. 

51 


52  STRUCTURAL  ENGINEERING 

Now  it  is  evident  that  the  part  cdB  of  the  beam  to  the  right  of  the  block 
must  exert  an  equal  force  S  (neglecting  the  weight  of  the  block  abed) 
upward  along  the  section  cd  upon  the  block  in  order  to  prevent  the  block 
from  being  pulled  downward  by  the  force  S  acting  downward  upon  it 
along  the  section  ab.  It  is  readily  seen  that  the  downward  force  S  acting 
along  the  section  ab  upon  the  block  and  the  equal  and  opposite  force  S 
acting  upward  along  the  section  cd  have  a  tendency  to  break  the  material 
in  the  block  off  vertically.  This  action  is  most  readily 
s  ^  comprehended  by  imagining  the  sections  ab  and  cd  to 

-^ WMk.     approach  each  other  until  the  block  abed  is  really  a 

single   layer   of   molecules.      Then   there   would   be   a 
force  S  acting  downward  along  ab  upon  the  left  side 


56  of  the  molecules  and  an  equal  and  opposite  force  S 

acting  upward  along  cd  upon  the  right  side  of  the 
molecules,  thus  tending  to  break  the  molecules  off  crosswise  instead  of 
tending  to  crush  or  pull  them  apart  as  in  the  case  of  simple  stress,  and 
the  stress  thus  produced  would  be  known  as  the  shearing  stress  on  the 
beam  at  that  point.  Whenever  forces  act  in  this  crosswise  manner,  in  any 
case,  the  stress  directly  produced  is  known  as  shearing  stress. 

In  speaking  of  the  shear  on  a  beam,  or  on  any  other  body,  we  refer 
to  it  as  being  x  pounds  along  a  certain  section  of  x  pounds  cut  by  an 
imaginary  plane;  but  what  we  really  mean  is  that  a  stress  of  x  pounds  is 
produced  upon  a  strip  of  material  infinitesimal  in  thickness  adjoining 
that  section. 

It  is  evident  that  in  the  above  case  the  shearing  stress  on  the  section 
ab  (really  the  shear  on  the  block  abed)  of  the  beam  is  equal  to  the  load  P 
and  to  the  weight  of  the  part  abA  of  the  beam.  The  part  of  the  shear  due 
to  the  load  P  is  the  same  for  all  vertical  sections  of  the  beam  between  A 
and.B,  while  the  part  due  to  the  weight  of  the  beam  will  evidently  be 
directly  proportional  to  the  length  of  the  part  abA.  Then,  if  w  be  the 
weight  of  the  beam  per  foot  of  length,  the  shear  at  any  vertical  section  x 
feet  from  the  end  A  will  be  S  =  P  +  wx.  Here  it  is  seen  that  the  shear  on 
any  vertical  section  of  the  above  beam  is  equal  to  the  algebraic  sum  of  the 
forces  between  the  section  and  the  end.  If  additional  forces  were  applied, 
they  would  likewise  be  included  in  the  summation,  provided  they  were  to 
the  left  of  the  section  considered.  As  the  forces  upon  the  part  of  the 
beam  to  the  right  of  the  block  abed  must  exert  an  upward  force  upon  the 
block  along  the  section  cd  equal  to  the  downward  force  along  ab  exerted 
by  the  forces  to  the  left,  it  is  evident  that  the  shear  on  the  block  is  equal  to 
the  algebraic  summation  of  the  forces  on  either  side  of  it.  So  is  the  case 
of  all  beams,  where  the  forces  are  applied  perpendicularly  to  the  longi- 
tudinal axis,  and  hence  we  have  in  general:  The  shear  on  any  section  of 
a  beam  perpendicular  to  its  longitudinal  axis  is  equal  to  the  algebraic 
summation  of  the  forces  on  either  side  of  the  section,  provided  all  of  the 
forces  act  perpendicularly  to  such  axis. 

This  is  readily  seen  to  be  true.  For,  let  AB  (Fig.  57)  represent  a 
beam  in  equilibrium  acted  upon  by  six  forces  PI  ...  P6,  no  reference 
being  made  as  to  which  are  reactions;  let  abed  be  an  imaginary  block 
cut  through  the  beam,  the  same  as  the  block  abed  shown  in  Fig.  56 :  It  is 
obvious  that  the  forces  PI  and  P2  would  cause  the  part  abA  of  the  beam 


THEOEETICAL  TREATMENT  OF  BEAMS  53 

to  exert  a  force  S  upon  the  block  along  ab  as  shown,  and  it  is  readily  seen 

that  this  force  S,  which  is  the  shear  on  the  block  (neglecting  the  weight  of 

the  beam)  would  be  equal  to  PI  +  P2,  that  is,  their  sum.    As  the  block  is 

in  equilibrium,  undoubtedly  there  must  be  an  equal  and  opposite  force  S 

exerted  upon  the  block  along  cd,  as  shown,  by  the  part  cdB  of  the  beam 

due  to  the  forces  acting  upon  that  part.     Then  we  have  P1+P£=+P6— 

P5+P4-P3  =  S;  that  is,  the  shear  on  the  block 

5.     i  ips  abed  is  equal  to  the  algebraic  summation  of  the 

old  J 1 —         forces  on  either  side  of  it.    Likewise,  the  shear  on 

8      any  section  between  P6  and  P5  is  equal  to  either 
P6     P6    or  P1+P2-P3  +  P4-P5;   between  P5   and 
FIg.  57  P4  it  is  equal  to  either  P6-P5  or  P1+P2-P3  + 

P4;  and  between  P4  and  P3  it  is  equal  to  either 

P6-P5+P4  or  P1+P2-P3.  This  means  that  the  shear  at  any  cross- 
section  of  any  beam  (as  stated  above)  is  equal  to  the  algebraic  summa- 
tion of  the  forces  on  either  side  of  the  section,  provided  the  forces  are 
perpendicular  to  the  beam. 

It  is  common  practice  to  assume  this  shearing  stress  to  be  uniformly 
distributed  over  the  cross-section  of  the  beam.  Then  according  to  this 
assumption,  the  shearing  stress  per  square  unit  on  any  cross-section  of 
the  beam  is  equal  to  the  shear  at  the  section  divided  by  the  area  of  the 
cross-section.  So  if  p  —  the  shearing  stress  in  pounds  per  square  inch  of 
cross-section,  S  =  the  total  shear  on  the  cross-section  in  pounds,  and  A  = 
the  area  of  the  cross-section  in  square  inches,  we  have  for  the  shearing 
stress  per  square  inch  the  general  formula 


The  assumption  that  the  shearing  stress  is  uniformly  distributed  over  the 
cross-section  is  not  absolutely  true,  as  will  be  shown  later;  however,  it 
meets  the  requirements  of  practical  designing  in  most  cases. 

53.  Bending  Stress  on  Beams. —  Imagine  for  the  time  being  that 
the  block  abed  (Fig.  56)  be  removed  and  suppose  instead  the  parts  abA 
and  dcB  of  the  beam  to  be  connected  by  a  hinge  placed  in  the  center  of 
the  cross-section  of  the  beam  as  shown  in  Fig.  58.  Now  it  is  evident  that 
this  hinge  could  be  so  constructed  that  it  would  prevent  the  part  abA  of 
the  beam  from  moving  down  vertically.  This  means 
that  the  hinge  would  resist  the  shearing  force  or 
"shear"  which  was  resisted  by  the  block  abed,  but 
it  is  evident  that  the  part  abA  would  now  fall  by 
rotating  downward  to  the  left  about  the  hinge.  So  it 
is  seen  that  the  part  abA  of  the  beam  has  a  tendency 
to  rotate  about  the  block  abed,  and  hence  will  be  subjected  to  stresses 
therefrom.  All  stresses  produced  in  this  manner  in  beams,  or  any  other 
bodies,  are  known  as  bending  stresses,  or  as  stresses  due  to  cross  bending. 

To  determine  the  bending  stress  in  the  block  abed  of  the  above  beam 
(Fig.  56),  let  us  assume  all  of  the  block  removed  except  a  thin  horizontal 
strip  ad  at  the  top  of  the  beam  and  an  equal  strip  be  at  the  bottom,  as 
shown  in  Fig.  59.  For  the  sake  of  simplicity  let  us  first  consider  the 
stress  on  these  strips  due  to  the  load  P  only.  The  load  P  would  cause  the 


54 


STRUCTURAL  ENGINEERING 


P 

d 

X 

\    h      Q 

i 

s 

Fig.   59 


part  abA  of  the  beam  to  exert  a  downward  vertical  force  s  upon  each  strip 
(see  Fig.  59)  which  must  be  resisted  by  an  upward  force  s  exerted  upon 
the  part  abA  in  turn  by  each  strip.  These  two  upward  forces  s  upon  the 
part  abA  and  the  load  P  form  a  couple  whose  moment  is  2sx  or  Px,  as 
2s  —  P.  This  couple  tends  to  rotate  the  part  abA  of  the  beam  counter 
clock-wise,  but  actual  rotation  is  prevented  by  the  strips  ad  and  be,  and 
consequently  each  will  be  subjected  to  a  stress.  Now  as  it  requires  a 
couple  to  balance  a  couple  (see  Art.  44)  the  stress  produced  upon  the  two 
strips  must  form  a  couple  and  the  stress  in  strip 
ad  must  act  to  the  right,  while  the  stress  in  the 
strip  be  must  act  to  the  left  upon  the  part  abA  of 
the  beam.  Let  F  be  the  stress  produced  on  each 
strip  and  let  h  be  the  vertical  distance  between 
their  centers  of  cross-section.  Then  we  have  Fh  = 
Px,  from  which  we  obtain  F=  (Px*)/h  as  the  stress 
on  each  strip.  So  far  our  problem  is  very  simple. 
But  suppose  we  add  two  other  strips,  mm  and  nn,  the  same  in  section  as 
the  two  just  considered  and  which  are  also  symmetrically  arranged  in 
reference  to  the  longitudinal  axis  of  the  beam  as  shown  in  Fig.  60.  Now 
it  is  evident  that  the  additional  strips  will  help  to  resist  the  moment  of 
the  couple  Px,  so  that  the  stress  F  on  the  two  strips  just  considered  will 
be  reduced.  Let  /  and  /'  now  be  the  stresses  on  the  strips,  and  h  and  h' 
their  lever  arms,  as  indicated  in  Fig.  60.  Then  we  have  Px  =  fh  +  f'h'. 
Here  it  is  seen  that  we  have  one  equation  and  two 
unknown  quantities,  /  and  /',  and  it  is  readily  seen 
that  for  each  additional  pair  of  such  strips  com- 
posing the  block  added,  another  unknown  force 
will  occur,  so  our  problem  is  not  as  simple  as  it 
first  appeared.  However,  we  get  out  of  our  diffi- 
culty by  resorting  to  Hooke's  Law  (see  Art.  32). 
It  is  seen  that  the  strips  at  the  top  of  the  beam  are 

in  tension  while  those  at  the  bottom  are  in  compression,  consequently  the 
strips  at  the  top  will  be  lengthened  while  those  at  the  bottom  will  be 
shortened,  and  the  block  abed,  which  is  rectangular  when  not  subjected  to 
bending  stress,  will  really  be  wedge-shaped  as  shown  in  Fig.  61.  As  the 
top  strips  are  in  tension  and  the  bottom  ones  in  compression,  it  is  evident 
that  there  must  be  a  horizontal  strip  of  fibers  somewhere  between  the  top 
and  bottom  of  the  beam  which  has  neither  tension  nor  compression;  that 

is,  no  stress  at  all.  Let  og  be  the 
location  of  this  strip  which  is  known 
as  the  neutral  plane  or  neutral  axis. 
As  all  the  strips  above  og  are  in  ten- 
sion and  all  below  are  in  compression, 
it  is  obvious  that  g  is  the  center  of 
tendency  of  rotation  of  the  part  abA 
of  the  beam.  Then,  evidently,  the 
distortion  of  the  strips  both  above  and  below  og  will  vary  directly  as  their 
distance  out  from  og.  Now  as  the  stress  on  any  strip,  according  to 
Hooke's  Law,  will  be  directly  proportional  to  its  distortion,  the  stress  per 
square  inch  on  any  strip  will  be  directly  proportional  to  its  distance  out 
from  ogt  and  hence  the  stress  on  the  strips  may  be  represented  by  the 


Fig.  GO 


Fig.   61 


THEOEETICAL  TREATMENT  OF  BEAMS  55 

arrows  enclosed  in  the  triangles  nkr  and  mpr  as  shown  in  Fig.  61,  where 
the  lines  kr  and  mr  have  the  same  slope  with  the  vertical. 

Let  /  be  the  stress  per  square  inch  on  any  horizontal  strip  of  fibers 
y  distance  out  from  og,  and  let  /'  be  the  stress  per  square  inch  on  a  strip 
out  unit  distance  from  og.  Then  we  have  the  proportion 

1  =  9  ••  f  •  f- 
from  which  we  get 

/'=-?• 

y 

Now  as  f/y  is  the  stress  on  a  strip  out  unit  distance  from  og,  the  stress  on 
a  strip  out  any  distance  z  from  og  would  be  z  times  f/y  or  (f/y)z,  and 
the  moment  of  it  about  g  would  be  s  times  this  or  (f/y)z2.  Now  /  was 
taken  as  so  many  pounds  per  square  inch,  but  as  the  stress  varies  con- 
tinuously from  og  outwardly,  there  will  not  be  a  square  inch  of  material 
anywhere  having  the  same  stress,  so  that  each  strip  must  really  be  con- 
sidered as  being  a  horizontal  element;  that  is,  a  mere  plane  of  fibers 
having  an  infinitesimal  area  of  cross-section  da.  For  the  sake  of  concep- 
tion we  may  say  that  da  is  1/1,000,000  of  a  square  inch.  Then  the  actual 
stress  or  force  out  z  distance  from  og  would  be  1/1,000,000  of  (f/y)z  or 
(ffy)zda  (da  =  dzb)  (see  section  Fig.  61),  and  the  moment  of  it  about  g 
would  be  z  times  that,  or  (f/y)z2da.  Now,  undoubtedly  the  summation  of 
these  moments  (f/y)z2da  about  g  for  all  of  the  strips  in  the  block  abed, 
which  we  can  express  as  (f/y)*%z2da,  must  be  equal  to  the  moment  of  the 
vertical  couple  Px.  So  we  have 


But  2z2da  =  I 

(see  Art.  49),  the  moment  of  inertia  of  the  cross-section  of  the  beam. 
Then  we  have  the  equation 

p*=li. 

y 

It  will  be  observed  very  readily  that  Px  is  simply  the  moment  of  the 
load  P  about  the  section  ab  (ab  being  considered  vertical,  and  practically 
it  is)  ;  that  is,  Px  is  the  measure  of  the  tendency  of  rotation  of  the  part 
abA  of  the  beam  about  the  section  ab  due  to  the  load  P.  Then  evidently 
the  moment  of  any  other  load  to  the  left  of  the  section  ab  about  the  section 
would  be  the  measure  of  the  tendency  of  rotation  of  the  part  abA  of  the 
beam  about  the  section  due  to  any  such  loads.  Then  if  ^Px  be  the  alge- 
braic summation  of  the  moments  of  any  number  of  loads  on  the  left  of  the 
section  about  the  section,  we  have  the  formula 


y 

from  which  the  stress  /  on  any  horizontal  element  of  the  block  abed  due  to 
the  loads  can  be  computed  by  substituting  for  y  the  distance  of  the  element 
out  from  og,  and  /  would  be  known  as  the  bending  stress  on  the  element, 
the  block  abed  being  considered  infinitesimal  in  length. 

Now  it  is  obvious  that  the  above  formula  will  apply  to  any  section 
of  the  above  beam  if  SPj?  be  taken  as  the  algebraic  summation  of  the 


56  STRUCTURAL  ENGINEERING 

moments  of  the  loads  to  the  left  of  that  section  about  the  section.  But  it 
is  readily  seen  that  the  ^Px  on  one  side  of  any  section  must  be  equal  and 
opposite  to  the  ^Px  on  the  other  side  in  order  that  the  beam  may  not 
rotate  about  the  section.  This  is  undoubtedly  true  in  the  case  of  any 
beam  whatever.  That  means  that  the  above  formula  applies  to  any 
section  of  any  beam  if  ^Px  be  taken  as  the  algebraic  summation  of  the 
moments  of  the  forces  (both  applied  forces  and  reactions)  on  either  side 
of  the  section  about  the  section.  This  summation  is  known  as  the  bend- 
ing moment  at  the  section  and  is  usually  represented  by  the  letter  M. 
Then  we  can  write  the  general  equation  as 


and  by  transposing  we  have 

(D), 

which  is  the  stress  on  any  horizontal  element  y  distance  from  the  neutral 
axis  either  above  or  below  it.  It  will  be  observed  that  the  neutral  axis 
extends  all  the  way  along  the  beam. 

It  is  seen  that  y  varies  from  zero  to  either  +e  or  -t  (Fig.  61)  and 
that  /  will  be  a  maximum  when  y  is  a  maximum,  as  M  and  /  are  constants 
for  each  specific  section.  So  if  y  be  taken  as  the  distance  to  the  farthest 
element  out  from  g  the  maximum  value  of  /  will  be  obtained.  This  is  what 
is  usually  desired,  and  for  that  reason  y  is  usually  spoken  of  as  the  dis- 
tance to  the  extreme  fiber  and  /  as  the  stress  on  the  extreme  fiber. 

The  summation  of  the  bending  stresses  above  the  neutral  axis  is 
always  equal  to  the  summation  of  the  bending  stresses  below  it,  but  the 
sums  of  their  moments  about  the  neutral  axis  are  not  equal  except  in  the 
case  of  symmetrical  section.  But  each  force  on  one  side  is  always  paired 
with  an  opposite  and  equal  force  on  the  other  side,  thus  forming  couples 
throughout  the  cross-section. 

54.     Reaction  on  Simple  Beams.— In  order  to  determine  the  shear- 
ing and  bending  stress  in  a  beam,  it  is  first  necessary  to  know  all  of  the 
external  forces  acting  upon  it.     The  applied  forces,  or  loads  as  they  are 
known,  are  usually  given,  but  the  reactions  are 

<J. j         usually  unknown  and  hence  have  to  be  deter- 

H -c- -J         mined  before  the  stresses  in  the  beam  can  be 

determined.  We  usually  resort  to  the  equations 
of  moments  to  determine  the  reactions  on  beams. 
However,  they  can  be  graphically  determined, 
as  will  be  shown  later. 


Fig.  02  Let  AB  (Fig.  62)  represent  a  beam  which 

is  supported  at  A  and  B  and  which  in  turn  sup- 

ports the  loads  PI,  P2,  P3,  and  P4.  The  reaction  R  at  A  due  to  the 
above  loads  can  be  obtained  by  taking  moments  about  B,  and  the  reaction 
wRl  at  B  can  be  obtained  by  taking  moments  about  A.  Thus,  taking 
moments  about  Bf  we  have 


THEOKETICAL  TKEATMENT  OF  BEAMS  57 

from  which  we  obtain 

„      aPl  +  bP2  +  cP3  +  dP4: 

-TT 

and  by  taking  moments  about  A  we  can  determine  R~L  in  the  same  manner. 
It  will  be  observed  that  by  taking  moments  about  the  line  of  action  of  one 
of  the  reactions  we  eliminate  that  reaction  from  the  equation  of  moments, 
thus  reducing  the  number  of  unknowns  in  the  equation  to  one.  The  sum 
of  the  moments  being  equal  to  zero  is,  in  accordance  with  the  laws  of 
equilibrium,  that  the  moments  of  the  external  forces  acting  upon  a  body  in 
equilibrium  about  every  point  must  be  equal  to  zero  (see  Art.  42).  The 
above  method  is  quite  general,  provided  the  loads  are  fully  known  and  the 
reactions  are  limited  to  two  in  number,  which  is  always  the  case  for 
simple  beams. 

55.  Graphical  Representation  of  Shear  on  Simple  Beams. — 
The  shear  on  a  beam  at  any  section  is  equal  to  the  algebraic  sum  of  the 
loads  aad  reaction  (as  stated  in  Art.  52),  summed  up  to  the  section,  begin- 
ning at  either  end.  For  example,  the  shear  at  every  section  between  the 
loads  P3  and  P2  (Fig.  62)  is  equal  to  either  R  - P4  - P3 or  R1-P1-P2; 
between  A  and  P4  it  is  equal  to  either  R  or  Rl  -PI  -P2-P3-P4; 
between  B  and  PI  it  is  equal  to  either  Rl  or  R  -P4-P3-P2-P1. 

This  shear  on  the  above  beam  can  be  graphically  represented  in  the 
following  manner: 

Draw  a  horizontal  line  ab  (Fig.  63)  equal  to  L,  the  length  of  the 
beam  (between  centers  of  end  bearings),  to  some  convenient  scale.  Then 
draw  the  lines  of  action  of  the  forces  in  their  relative  positions  as  shown 
at  1,  2,  3,  and  4.  Then  beginning  at  one  end, 
say,  at  a,  draw  ac  =  R.  Then  through  c  and 
parallel  to  ab  draw  cd  and  from  d  lay  off  de  = 
P4.  Then  from  e  draw  ef  and  from  /  lay  off 
fg  =  P3,  and  so  on,  thus  obtaining  the  zigzag  line 
cde  .  .  .  n.  Any  ordinate  as  rr  from  the  line 
ab  to  the  zigzag  line  represents  the  shear  on 
Fig.  63  the  beam  at  that  point. 

The  value  of  the  shear  on  any  vertical  sec- 
tion o-f  a  beam  will  depend  upon  the  weight  of  the  load  or  loads  it  sup- 
ports and  upon  the  position  they  occupy  in  reference  to  the  section. 
Loads  are  known  as  dead  load  and  live  load. 

The  dead  load  upon  a  beam  is  the  weight  it  supports  that  is  fixed  in 
position,  while  the  live  load  is  the  weight  it  supports  that  is  not  fixed  in 
position  but  may  move  to  any  position  upon  it.  The  dead  load  of  struc- 
tures usually  consists  of  the  weight  of  the  structure  itself,  while  the  live 
load  consists  of  loads  which  it  supports  but  which  at  the  same  time  move 
over  it,  as  a  locomotive,  train  of  cars,  wagons,  etc.  The  dead  load  is 
usually  considered  as  a  uniformly  distributed  load.  The  live  load  is 
sometimes  considered  as  a  uniformly  distributed  load,  but  more  often  as 
concentrated  loads. 

If  the  load  be  a  uniformly  distributed  dead  load,  each  reaction  will 
be  R  —  wL/2,  where  L  is  the  length  of  the  span  and  to  the  weight  of  the 
dead  load  per  foot  of  span,  and  the  shear  out  any  distance  z  from  either 
end  will  be  S  =  R-zw=wL/2-zw.  It  is  seen  that  when  z  =  L/2  the 


P4- 


P2 


58 


STRUCTURAL  ENGINEERING 


shear  is  equal  to  zero,  and  when  2  =  0  the  shear  is  equal  to  wL/2.  This 
shows  that  the  shear  due  to  such  a  load  is  a  maximum  at  the  ends  and 
zero  at  the  center. 

The  above  equation  for  shear,  S  =  wL/2-zw,  is  an  equation  to  a 
straight  line,  so  evidently  the  shear  can  be  represented  graphically  by  the 
ordinates  to  a  straight  line,  as  acb  (Fig.  64),  where  AB  represents  the 

length  of  span  and  aA  and  bB  are  equal  to 
tvL/2.  The  shear,  beginning  at  A,  decreases 
until  c,  the  center  of  the  span,  is  reached, 
where  it  is  zero ;  then  beyond  that  point,  still 
summing  up  to  the  right,  the  shear  will  be 
minus,  increasing  toward  B,  being  equal  to 
—wL/2  at  B.  So  it  is  seen  that  the  ordinates 
Fig.  64  to  the  line  acb  truly  represent  shear  for  dead 

load    uniformly    distributed.      For    example, 

the  ordinate  fe  represents  the  shear  at  /,  which  is  positive,  while  kh  rep- 
resents the  negative  shear  at  k.  The  sign  of  the  shear  has  no  practical 
significance  other  than  algebraic. 

In  case  of  uniform  live  load  moving  over  a  beam,  the  shear  at  any 
section  will  be  a  maximum  when  the  load  just  extends  from  one  of  the 
ends  up  to  the  section.  For  example,  suppose  a  live  load  of  p  pounds  per 
foot  extends  from  B  up  to  some  section  0,  a  point  x  feet  from  B.  Then 
the  shear  at  0  is  equal  to  the  reaction  at  A,  which  expressed  in  terms  of 
the  load  is  (px)x/2  +  L  =  px2/2L.  It  is  evident  that  the  shear  produced 
at  section  0,  as  the  load  moves  from  B  to  0,  keeps  increasing  as  the  load 
approaches  0,  as  it  is  equal  to  the  reaction  at  A,  which  keeps  increasing 
as  the  load  moves  from  B  to  0.  Now  if  the  load  extends  past  0  to  any 
section  m,  the  reaction  at  A  will  be  increased  a  certain  amount  by  the 
additional  load  py,  but  in  obtaining  the  shear  at  0  we  subtract  all  of  the 
load  py  from  the  reaction  at  A,  and  as  all  of  the  additional  load  py  is  not, 
as  we  say,  transmitted  to  the  end  A,  it  is  evident  that  the  load  extending 
beyond  0  will  diminish  the  shear  at  that  section.  Therefore  the  above 
statement  is  shown  to  be  true,  and  the  equation  S  =  px2/2L  is  the  general 
equation  for  the  maximum  shear  produced  on  a  beam  by  a  uniform  live 
load  moving  over  it.  The  equation  is  that  of  a  parabola.  Then,  evidently, 
if  we  construct  a  parabola  as  auB  so  that  Aa  is  equal  to  pL/2,  the  shear 
at  any  section  will  be  represented  by  the  ordinate  to  the  parabola  at  that 
section.  For  example,  the  shear  at  t  is  represented  by  the  ordinate  ut. 
The  shear  curve  auB  (Fig.  64),  as  stated  above,  is  an  absolute 
parabola  for  a  uniform  live  load,  but  for  such  loads  as  the  wheels  of  a 
locomotive  it  would  be  a  curve  made  up  of  a  series  of  straight  lines, 
however,  if  the  loads  are  nearly  equal  in  weight  and  quite  uniform  in 
spacing,  the  curve  will  approach  a  parabola,  and  in  most  cases  it  is  near 
enough  to  be  considered  one  for  practical  purposes. 

56.  Graphical  Representation  of  Bending  Moments  on  Simple 
Beams. — 

Case  I.  When  the  load  is  uniformly  distributed  over  the  entire 
length. 

Let  AB  (Fig.  65)  represent  a  beam  supported  at  A  and  B  which 
in  turn  supports  a  uniform  load  of  w  pounds  per  lineal  foot  of  length. 
The  bending  moment  at  any  section  ss,  x  feet  from  A  due  to  this  load  is 


THEORETICAL  TREATMENT  OF  BEAMS 


But 

so  we  have 


R  = 


wLx 


was* 


for  the  bending  moment  at  any  section  of  the  beam.     When  x  =  L/2,  we 
have 

wL2     wL2 


from  which  we  obtain 


(E), 


which  is  a  formula  of  great  practical  value  as  it  expresses  the  moment  at 
the  center  of  any  simple  beam  uniformly  loaded,  which  is  the  maximum 
moment  that  can  occur  under  such  loading,  and  usually  this  is  what  we 
desire  to  know. 

The  above  formula,  M  =  wLx/%  -  wxz/2)  is  an  equation  to  a 
parabola.  Then,  evidently,  the  bending  moments  at  the  different  sections 
along  the  beam  vary  as  the  ordinates  to  a  par- 
abola. It  is  seen  from  the  equation  that  M  — 0 
when  x  =  Q,  and  also  when  x  —  L,;  and  when  x 
=  L/2,  M  =  wL2/S.  Then  if  we  draw  CO  equal  to 
wL2/S  to  scale,  and  perpendicular  to  the  beam 
at  its  center  and  pass  a  parabola  through  A,  O, 
and  B  as  shown  in  Fig.  65,  any  ordinate  as  rr 
will  represent  the  bending  moment  on  the  beam 


Fig.  65 


When  the  loads  are  concentrated  at  different  points  along 


at  that  point. 

Case  II. 
the  beam. 

Let  AB  (Fig.  66)  represent  a  beam  supported  at  A  and  B  which 
in  turn  supports  the  loads  PI  .  .  .  P5,  as  shown.  The  bending  moment 
anywhere  between  A  and  P5,  due  to  these  loads,  is  M  =  Rx,  which  is  an 
equation  to  a  straight  line.  When  ,r  =  0,  M  =  0,  and  when  x  —  a,  M  =  Ra. 
Then,  if  we  draw  cd  equal  to  Ra,  to  scale,  and 
join  A  and  d,  any  ordinate  as  ss  to  the  line  Ad 
will  represent  the  bending  moment  on  the  beam 
at  that  point.  The  bending  moment  anywhere 
between  Po  and  P4  is  M  =  .R(a  +  ,r')  —  P5#', 
which  is  also  an  equation  to  a  straight  line.  When 
#'  =  0,  M  =  Ra,  so,  evidently,  one  point  in  the 
line  represented  by  the  equation  M  —  R  («  +  #') 

-  PS*'  is  d,  as  cd  =  Ra.  When  x'  =  b,  M  =  R  (a  +  6)  -  P56.  So  if  we  draw 
ef  equal  to  R(a  +  b)-P5b  to  scale,  using  the  same  scale  as  in  the  case 
of  cd,  we  obtain  the  line  df,  the  ordinates  to  which  evidently  represent 
the  bending  moments  between  P5  and  P4.  Now  it  is  evident  that  the 
lines  fg,  gh,  hk,  and  TcB  can  be  similarly  constructed,  the  ordinates  to 


60 


STRUCTURAL  ENGINEERING 


which  represent  the  bending  moments  between  P4  and  P3,  P3  and  P2, 
P2  and  PI,  PI  and  B,  respectively.  So  it  is  seen  that  the  bending 
moments  on  a  simple  beam  due  to  any  number  of  concentrated  loads 
can  be  graphically  represented  by  the  ordinates  to  a  series  of  straight 
lines  forming  a  closed  polygon  with  the  ends  of  the  beam.  The  nearer 
the  loads  approach  a  uniform  load,  the  nearer  this  polygon  approaches 
a  parabola. 

It  is  important  to  note  that  the  maximum  moment  always  occurs 
under  a  load.  This  is  readily  seen,  for  the  maximum  ordinate  to  any 
segment  of  the  above  polygon  is  under  a  load  and  hence  the  maximum 
of  them  all  will  be  under  a  load. 

57.  Graphical  Representation  of  the  Bending  Moments  and 
Shears  on  Cantilever  Beams. — 

Case  I.      When  the  load  is  uniformly  distributed 
over  the  entire  length. 

Let  AB  (Fig.  67)  represent  a  cantilever  beam 
supporting  a  uniform  load  of  w  pounds  per  foot  of 
length.  The  bending  moment  at  any  section  ss,  x  feet 
from  A,  is  M=  (wx)x/%  =  wx2/2,  which  is  an  equation 
to  a  parabola.  Making  x  =  L,  we  have  M  =  wL2/%, 
which  is  the  bending  moment  at  B.  Then  if  we  draw 
the  horizontal  line  ab  =  L,  and  the  vertical  line  cb  — 
wL2/2,  to  scale,  and  construct  the  parabola  ac,  the 
ordinates  between  the  line  ab  and  the  curve  ac  will 
graphically  represent  the  bending  moments  on  the  beam  due  to  the  above 
load.  For  example,  the  moment  at  the  section  ss  is  represented  by  the 
ordinate  rr. 

The  shear  at  any  section  ss  is  S  =  wx,  which  is   an  equation  to  a 
straight  line.     When  x  =  0,  S  =  0,  and  when  x  —  L,t  S  =  wL,  which  is  the 
shear  at  B.     Then,  if  we  draw  bo  equal  to 
•wL,  to  scale,  and  join  a  and  o,  the  shear  at 
any  section  in  the  beam  is  represented  by 
the   ordinate   between   the   line   ab   and   ao 
immediately  under  the  section.     For  exam- 
ple, the  shear  at  the  section  ss  is   repre- 
sented by  the  ordinate  er. 

Case  II.  When  the  loads  are  concen- 
trated at  different  points  along  the  beam. 
Let  AB  (Fig.  68)  represent  a  canti- 
lever beam  supporting  the  concentrated 
loads  PI  .  .  .  P4  as  shown.  As  a  general 
example,  the  bending  moment  due  to  the 
above  loads  at  a  section  as  ss,  expressed 
analytically,  is  M  =  nPl  +  bP2  +  tP3.  If  we 
proceed  in  the  same  manner  as  in  the  case 
of  a  simple  beam,  making  the  horizontal 
line  ab  =  L,  and  the  verticals  wx,  w'x', 
etc.,  under  each  load,  equal  to  the  bending  Fig.  cs 

moment    about    each,    respectively,    making 

bv  equal  to  the  moment  at  B,  and  drawing  the  broken  line  aww'w"v, 
we  have  the  bending  moments  on  the  beam  graphically  represented  by 


'THEORETICAL  TREATMENT  OF  BEAMS 


61 


the  ordinates  between  the  line  ab  and  this  broken  line.  For  example, 
the  moment  at  the  section  ss  is  represented  by  the  ordinate  ro.  The 
shear  at  the  section  ss  is  S  =  P1+P2  +  P3,  expressed  analytically.  Con- 
structing the  zigzag  line  acd . . .  m,  the  same  as  in  the  case  of  simple 
beams  (Art.  55).,  we  have  the  shear  at  any  section  represented  by  the 
ordinate  between  the  line  ab  and  this  zigzag  line. 

58.  Graphical  Construction  of  a  Parabola. —  The  following  method 
of  constructing  a  parabola  will  be  found  to  be  quite  convenient.  It  really 
consists  in  passing  a  parabola  through  two  points,  when  one  of  the  points 
is  taken  at  the  vertex.  Let  A  and  B  (Fig.  69)  be  any  two  given  points 
and  let  A  be  at  the  vertex  of  the  parabola  desired.  Draw  the  line  XX 
through  A  as  the  X-axis  of  the  parabola.  Then  draw  AC  perpendicular 


to  XX  and  BC  perpendicular  to  AC.  Divide  AC  and  BC  into  an  equal 
number  of  equal  parts,  say,  six,  as  shown.  Then  from  A  draw  the  radial 
lines  \-A,  2-A,  etc.,  and  where  the  radial  line  \-A  intersects  the  vertical 
aa  is  one  point  on  the  curve  of  the  required  parabola,  and  where  the  next 
radial  line  2-A  intersects  the  next  vertical  bb  is  another  point  on  the 
curve,  and  so  on.  The  other  side  AB'  of  the  parabola  can  be  constructed 
by  laying  off  AC' -AC  and  dividing  it  into  the  same  number  of  equal 
parts  as  AC  and  drawing  verticals  as  shown  and  then  projecting  the 
corresponding  points  over  from  the  curve  AB  just  constructed,  or  by 
dividing  B'C'  into  the  same  number  of  equal  parts  as  AC'  and  proceeding 
as  explained  for  the  other  side  of  the  curve. 

59.  Relation  of  Shear  to  Bending  Moment.— Let  AB  (Fig.  70) 
represent  a  simple  rectangular  wooden  beam  supported  at  A  and  B  and 
let  P  be  a  load  which  the  beam  supports  at  mid-span,  and  let  R  and  R\ 
be  the  reactions  (which  will  be  equal)  at  A  and  B,  respectively,  due 
to  this  load  P.  Suppose  the  reactions  applied  exactly  at  the  ends  of 
the  beams  as  indicated,  which  is  not  at  all  an  imaginary  case,  and  imagine 
the  beam  divided  up  into  short  rectangular  blocks  as  shown,  by  imaginary 
planes  passing  perpendicularly  to  the  longitudinal  axis  of  the  beam.  Let 
A,r  be  the  length  of  each  block.  The  reaction  R  at  A  applied  along  the 
end  of  the  first  block,  beginning  at  A,  tends  to  move  that  block  upward. 
If  the  block  does  not  move  undoubtedly  the  second  block  prevents  it  by 
exerting  an  equal  and  downward  force  R  where  the  two  blocks  join  as 
indicated.  As  the  first  block  is  held  entirely  by  the  second  block  it  is 
obvious  that  the  first  block  will  exert  an  upward  force  upon  the  second 
block  where  the  two  blocks  join,  and  if  the  second  block  does  not  move 
upward  undoubtedly  the  third  block  prevents  it  by  exerting  a  downward 


62 


STRUCTURAL  ENGINEERING 


Fig.   70 


force  /2  where  the  second  and  third  blocks  join,  and  hence  it  is  seen  that 
the  blocks  to  the  left  of  the  load  resist  each  other  in  transmitting  the 
shear,  each  thereby  receiving  an  upward  force  R  upon  one  end  and  an 

equal  downward  force  upon  the  other  end 
as  is  indicated.  The  same  is  true  of  the 
blocks  to  the  right  of  the  load  P  except 
the  forces  are  reversed  in  order.  These 
vertical  forces  not  only  tend  to  shear  the 
blocks  off  vertically  but  tend  to  rotate  them 
as  well,  as  the  two  forces  on  each  block 
form  a  vertical  couple. 

Now  beginning  at  A  and  considering  the  blocks  to  the  left  of  P  only, 
the  bending  moment  at  the  right  end  of  the  first  block  is  RAx ;  at  the  right 
end  of  the  second  block  it  is  R(2kx)  ;  and  at  the  right  end  of  the  third 
block  it  is  .R(3A,r)  ;  and  so  on,  being  R(nAx)  at  the  right  end  of  the  nth 
block  from  A,  which  is  seen  at  a  glance  to  be  nothing  more  than  the 
summation  of  the  moments  of  the  vertical  couples  on  the  blocks  beginning 
at  A.  This  means  that  the  bending  moment  beginning  at  the  end  A 
increases  toward  the  load  P  by  one  Rkx  at  each  block.  For  example,  the 
bending  moment  at  kk  is  one  Rkx  greater  than  it  is  at  hh.  So,  evidently, 
R&x  is  the  increment  of  the  bending  moment  to  the  left  of  the  load  P  in 
the  case  of  the  above  beam.  By  the  same  reasoning  we  have  Rl&x  as  the 
increment  of  the  bending  moment  to  the  right  of  the  load  P. 

It  will  be  observed  that  all  of  the  blocks  to  the  left  of  the  load  P 
tend  to  rotate  clock-wise  while  all  of  the  blocks  to  the  right  of  P  tend  to 
rotate  in  the  opposite  direction  or  counter  clock-wise,  so  that  the  sign  of 
the  increments  of  the  bending  moment  on  one  side  will  be  different  from 
those  on  the  other  side  of  the  load.  The  bending  moment  at  any  point 
to  the  left  of  the  load  P  is  equal  to  the  sum  of  the  increments  R&x 
between  A  and  the  point  and  the  bending  moment  at  any  point  to  the 
right  of  the  load  is  equal  to  the  sum  of  the  increments  R\kx  between  B 
and  the  point,  although  the  bending  moment  at  any  point  is  equal  to  the 
sum  of  the  increments  summed  up  algebraically  from  either  end  of  the 
beam.  For  example,  the  bending  moment  at  gm  is  equal  to  the  sum  of 
the  increments  to  the  right  of  gm  or  to  the  sum  of  the  increments  to  the 
left  of  P  minus  the  sum  of  the  increments  between  P  and  gm,  as  the  signs 
are  different. 

In  the  case  shown  in  Fig.  70  the  increments  of  the  bending  moment 
are  all  equal,  but  suppose  we  consider  the  case  shown  in  Fig.  71  where 
R  and  Rl  are  the  reactions  at  A  and  B,  respectively,  due  to  the  three  loads 
PI,  P2,  and  P3.  Here  it  is  readily  seen 
that  the  increment  of  the  bending  moment 
between  A  and  PI  due  to  PI,  P2,  and  P3 
is  flAx;  between  PI  and  P2  it  is  (R-Pl) 
Ax;  between  P2  and  P3  it  is  (R-P1-P2) 
Ax;  and  between  P3  and  B  it  is  (R -PI 
-P2-P3)Ax.  That  is,  the  increment  of 
the  bending  moment  at  any  point  is  equal  to  the  shear  at  that  point  multi- 
plied by  Ax.  It  will  be  observed  that  this  was  true  in  the  case  shown  in 
Fig.  70,  for  R  is  the  shear  to  the  left  of  P  and  Rl  the  shear  to  the  right 
of  P,  which  are  equal  in  that  case.  Now,  the  vertical  couples  on  any  such 


Fig.  71 


THEOKETICAL  TEEATMENT  OF  BEAMS  63 

blocks  or  strips,  as  here  considered,  in  any  beam,  could  not  be  anything 
other  than  the  vertical  shear  couples  on  them,  so  we  have,  in  general, 


for  the  increment  of  the  bending  moment,  where  S  is  the  shear  on  the 
block  in  question  and  A,r  its  length.  Now,  A.r  could  have  any  practical 
value  but  the  ultimate  increment  will  be  when  A,r  is  an  infinitesimal. 
Then  assigning  the  smallest  possible  value  to  A,r  making  it  an  infinitesimal 
dx,  we  obtain 

dM  =  Sdx 

as  the  ultimate  increment  of  the  bending  moment,  as  it  is  the  smallest 
increment  possible.  That  is,  the  differential  of  the  bending  moment  at 
any  section  of  any  beam  is  equal  to  the  vertical  shear  couple  on  an 
imaginary  vertical  block  of  infinitesimal  length.  The  preceding  formula 
is  usually  written 


• 


Expressing  this  in  words,  we  say  that  the  first  derivative  of  the  bending 
moment  in  reference  to  x  is  equal  to  the  shear,  where  x  is  the  distance 
from  the  end  of  the  beam  to  the  section  considered. 

The  above  equation  (F)  is  sometimes  quite  useful,  as  the  determining 
of  the  shear  from  the  bending  moment  is  sometimes  desirable.  The 
equation,  or  formula,  is  very  readily  applied,  as  is  seen  from  the  follow- 
ing, although  its  true  worth  is  more  readily  recognized  in  more  compli- 
cated cases. 

Considering  both  the  load  P  and  the  weight  of  the  beam  acting  upon 
the  beam  shown  in  Fig.  70,  the  bending  moment  at  any  section  x  distance 
from  the  end  A  is 

wx2 
M  =  Rx- 

10 

where  w  represents  the  weight  of  the  beam  per  foot.  Differentiating 
both  sides  of  the  equation  and  dividing  through  by  dx  we  have 

dM 

-j-^R-wx, 
dx 

which,  as  is  readily  seen,  is  the  shear  at  any  cross-section  x  distance  from 
the  end  A  and  to  the  left  of  the  load. 

The  bending  moment  at  any  section  x  distance  from  the  end  of  a 
simple  beam  having  a  length  L  and  supporting  a  uniform  load  of  w  pounds 
per  lineal  foot  is 

wLx     wx2  • 

M=—  -~r 

(see  Art.  56,  Case  I).     Differentiating  both  sides  of  this  equation, 
dividing  through  by  dx,  we  have 

dM     wL 

—  =  -  —  wx, 
dx        2 


64  STRUCTURAL  ENGINEERING 

which,  as  is  readily  seen,  is  the  general  expression  for  the  shear  at  any 
section  of  the  beam.  If  ,r  =  0,  the  shear  is  equal  to  wL/2,  and  if  x  —  I^y  it 
becomes  -(wL/2),  each  of  which  is  recognized  as  an  end  shear  or  reac- 
tion. If  x  =  L/2  the  shear  is  equal  to  0,  all  of  which  goes  to  show  how 
readily  Formula  (F)  can  be  applied,  even  for  ordinary  cases. 

60.  Increment  of  the  Bending  Stress.  —  Let  ab  (Fig.  70)  represent 
a  very  thin  horizontal  strip  through  the  beam  y  distance  above  the  neutral 
axis  oo.  The  end  block  at  A,  owing  to  its  tendency  to  rotate  clock-  wise, 
due  to  the  vertical  shear  couple  Rkx,  will  exert  a  force  to  the  right  upon 
all  horizontal  elements  in  the  second  block  above  the  neutral  axis  and  a 
force  to  the  left  upon  all  such  elements  in  the  second  block  below  the 
neutral  axis.  Then  undoubtedly  the  first  block  will  exert  a  force  to  the 
right  upon  the  strip  ab.  Let  A/  be  this  force.  The  end  block  at  B  has 
the  same  tendency  to  rotate  as  the  end  block  at  A  but  in  the  opposite 
direction,  and  hence  the  end  block  at  B  will  exert  a  force  A/  to  the  left 
upon  the  strip  ab.  Then  the  strip  ab  will  have  a  compressive  stress  of 
A/  in  it  from  c  to  d  due  to  the  end  blocks.  The  second  block  from  the 
end  A  having  the  same  tendency  to  rotate  as  the  end  block  will  likewise 
exert  a  force  A/  to  the  right  upon  the  strip  ab  and  the  second  block  from 
the  end  B  will  exert  a  force  A/  to  the  left  upon  the  strip  ab,  and  hence  the 
strip  ab  will  have  a  compressive  stress  of  2A/  from  e  to  /  due  to  the  first 
and  second  blocks  from  the  ends.  The  third  block  from  the  end  A  will 
exert  a  force  A/  to  the  right  upon  the  strip  ab  and  the  third  block  from  B 
will  exert  a  force  A/  in  the  opposite  direction  upon  the  strip  ab  and  hence 
the  strip  ab  will  have  a  compressive  stress  of  3A/  from  I  to  g  due  to  the 
first,  second,  and  third  blocks  from  the  ends.  Thus  it  is  seen  that  the 
stress  in  the  strip  ab  is  increased  at  each  block  by  one  A/  as  we  pass  from 
either  end  toward  the  load  P  where  the  stress  is  a  maximum.  So, 
evidently,  A/  is  the  increment  of  the  bending  stress  in  the  strip  ab.  As 
the  strip  ab  could  be  any  horizontal  element  out  any  distance  y  from  the 
neutral  axis,  either  above  or  below,  it  is  evident  that  the  bending  stress  on 
any  horizontal  element  in  the  beam  will  vary  by  such  increments  as  A/ 
the  same  as  in  the  case  of  the  strip  ab;  of  course,  their  real  values  will  be 
different  owing  to  their  distance  out  from  the  neutral  axis  being  different. 
For  the  value  of  A/  at  any  block  to  the  left  of  the  load  P  (Fig.  70)  we 
have 


and  for  its  value  to  the  right  we  have 


where  I  is  the  moment  of  inertia  of  the  cross-section  of  the  beam  in 
reference  to  the  horizontal  gravity  axis  of  the  beam;  y  the  distance  out 
from  the  neutral  axis  to  the  strip  or  element  considered;  and  Rkx  and 
/21A,r  the  vertical  couple  in  each  respective  case.  But  as  the  vertical 
couple  on  any  imaginary  vertical  block  or  strip  in  any  beam  whatever  is 
simply  the  shear  on  the  block  multiplied  by  its  length,  we  can  write  the 
general  expression  for  the  increment  of  the  bending  stress  at  any  point  in 


THEOEETICAL  TEEATMENT  OF  BEAMS  G5 

any  beam  as 

A/=  — j-^- (a). 

Then  beginning  at  either  end  of  a  beam  we  can  obtain  the  stress  on  any 
horizontal  element  x  distance  from  the  end  by  summing  up  the  stress 
increments  as  expressed  by  (a).  This  would  give  us 


But  SA/  =  /  and  ^S^x-M,  the  bending  moment  (see  Art.  59)  at  a  point  x 
distance  from  the  end,  when  A#  becomes  an  infinitesimal.  Then  substi- 
tuting /  for  2A/  and  M  for  2$A,r  in  the  last  equation  we  have 

My 

r~   I  > 
which  is  Formula  (D),  Art.  53. 

Owing  to  the  shear  being  constant  the  increments  of  the  bending 
stress  in  the  beam  shown  in  Fig.  70  are  constant  throughout  for  each 
horizontal  element  and  are  equal  to 


But  in  such  cases,  as  shown  in  Fig.  71,  they  will  be  different  owing  to  the 
variation  of  the  shear.  The  increment  between  A  and  the  load  PI,  in 
that  case,  due  to  the  loads  PI,  P2,  and  P3  is  expressed  as 

(flAg)y. 

/ 
between  PI  and  P2  as 

(fl-Pl)Asy. 

I 
between  P2  and  P3  as 

(fl-Pl-P2)A*y  , 

A/-  -— 

and  between  P3  and  B  as 

(fl-P'l-P2-P3)Aary  (JRlAjQy  § 

It  is  seen  from  the  above  that  the  increment  of  the  bending  stress 
varies  directly  as  the  shear  and  hence  in  most  cases  the  greatest  increment 
will  be  at  the  ends  of  a  beam,  especially  when  the  weight  of  the  beam  is 
considered. 

61.  Horizontal  Shearing  Stress  in  Beams.—  Let  AB  (Fig.  72) 
represent  a  portion  of  the  same  beam  shown  in  Fig.  56  (Art.  52),  which 
is  here  drawn  so  as  to  show  the  imaginary  block  abed  to  a  large  scale. 
Conceive  of  the  original  imaginary  block  abed  sub-divided  into  smaller 
imaginary  blocks  as  egkh,  gmnk,  etc.  We  can  now  conceive  of  the  shear- 
ing force  exerted  upon  the  block  abed  along  ab  and  cd  as  being  distributed 
to  each  of  the  smaller  blocks,  whereby  each  is  seen  to  have  a  vertical 
couple  which  we  can  conceive  of  as  being  an  increment  couple  of  the 


66 


STRUCTURAL  ENGINEERING 


vertical  couple  acting  upon  the  block  abed  as  a  whole.  Now,  evidently, 
each  of  the  smaller  blocks  has  a  tendency  to  rotate  owing  to  this  vertical 
couple,  and  if  rotation  does  not  take  place,  evidently  it  is  prevented  by 
the  material  joining  the  blocks  horizontally.  Then  the  forces  resisting  the 
vertical  couple  on  each  of  the  smaller  blocks  must  act  through  this  mate- 
rial, forming  a  horizontal  couple  on  each  block  as  shown,  which  must 
necessarily  be  equivalent  and  opposite  to  the  vertical  couple  that  it  resists. 
Now,  it  is  readily  seen  that  these  horizontal  couples  tend  to  shear  the 
smaller  blocks  off  horizontally  the  same  as  the  vertical  couples  tend  to 
shear  the  same  blocks  off  vertically.  If  a  block  be  square^  it  is  readily 
seen  that  the  vertical  and  horizontal  couples  on  it  will  not  only  be 
equivalent,  but  will  be  equal  and  opposite,  which  means  that  each  of  the 
forces  in  the  horizontal  couple  is  equal  to  each  of  the  forces  in  the  vertical 
couple;  that  is,  the  four  forces  acting  upon  the  block  are  equal.  This 
means  that  the  horizontal  shear  on  the  smaller  blocks  is  equal  to  the 
vertical  shear  on  the  same.  But  if  the  smaller  blocks  are  considered 

O    d 


e 

~=^~ 

*       1 

9 

-*s- 

* 

m 

1  — 

' 

b  c 

Fig.  72 

m      o 


Fig.  73 


shorter  or  longer  in  one  direction  than  in  the  other,  the  unit  shear  is  not 
changed.  The  larger  sides  having  the  greater  area  will  simply  have  a 
greater  force  acting,  but  the  force  per  square  unit  will  be  just  the  same, 
as  our  mentally  changing  the  block  will  not  alter  the  unit  shear.  As  far 
as  the  two  couples  on  any  block  remaining  equivalent  is  concerned,  it  is 
evident  they  will,  for  as  the  force  on  one  side  is  increased,  the  lever  arm 
of  the  other  is  increased,  so  that  the  moments  of  the  couples  are  con- 
tinually equivalent.  So  it  really  does  not  matter  what  shape  our  fancy 
molds  the  imaginary  parts,  for  it  remains  evident  that  the  horizontal  and 
vertical  shear  on  any  particle  in  any  beam  are  equal.  Then,  if  we  know 
the  horizontal  shear  on  any  particle,  we  know  the  vertical,  as  the  two  are 
equal,  and  vice  versa.  The  determining  of  the  total  vertical  shear  at  any 
vertical  section  of  a  beam  is  an  easy  matter,  but  just  what  the  intensity 
of  it  is  on  any  particular  particle  is  another  thing.  What  we  really  do  is 
to  find  the  horizontal  shear  and  consider  the  vertical  shear  equal  to  the 
same.  In  order  to  do  this,  let  abed,  Fig.  73,  represent  the  imaginary  block 
abed  of  Fig.  56,  as  an  independent  body  where  the  other  parts  of  the  beam 
are  replaced  by  the  forces  which  those  parts  exert  upon  the  block.  Let 
go  be  the  neutral  plane.  The  part  of  the  beam  to  the  left  of  the  block 
will  exert  upon  it  the  bending  stresses  represented  by  the  arrows  in 
triangles  agm  and  bgm',  and  also  the  vertical  shearing  force  S  along  ab, 
while  the  part  of  the  beam  to  the  right  will  exert  upon  it  the  forces 


THEORETICAL  TREATMENT  OF  BEAMS  <J7 

represented  by  the  arrows  in  the  triangles  dok  and  cok' ',  which  resist  the 
bending  stresses  represented  by  the  arrows  in  the  triangles  agm  and  bgm' ', 
and  hence  are  equal  and  opposite  to  them,  and  also  the  shearing  force  S 
along  dc,  which  is  equal  and  opposite  to  the  force  S  acting  along  ab,  and 
also  the  bending  stress- increments  represented  by  .the  arrows  in  the 
triangles  don  and  con'. 

The  bending  stresses  represented  by  the  arrows  in  triangles  agm  and 
bgm'  are  transmitted  directly  through  the  block  and  are  balanced  directly 
by  the  forces  represented  by  the  arrows  in  the  triangles  dok  and  cok' ,  so, 
evidently,  none  of  these  forces  causes  horizontal  shear  on  the  block  and 
can  be  disregarded.  The  force  S  acting  along  ab  and. the  equal  and 
opposite  force  S  acting  along  dc  and  the  bending  stress  increments  repre- 
sented by  the  arrows  in  the  triangles  don  and  con'  are  the  only  forces 
remaining.  But  as  each  of  the  forces  S  acts  vertically,  it  is  evident  that 
the  horizontal  shear  in  the  block  is  due  entirely  to  the  forces  represented 
by  the  arrows  in  the  triangles  don  and  con' ;  that  is,  to  the  increments  of 
the  bending  stresses  on  the  block,  and  hence  it  remains  for  us  to  consider 
these  forces  only. 

Now,  evidently,  the  horizontal  shear  on  any  element  as  ee  of  the 
block  abed  is  equal  to  the  algebraic  summation  of  these  increment  forces 
on  either  side  of  ee,  the  same  as  in  the  case  of  any  body.  Therefore,  the 
horizontal  shear  at  ee  is  equal  to  the  sum  of  these  forces  between  d  and  e, 
or  between  e  and  c. 

The  moment  of  these  increment  forces  is  equal  to  Skx,  which  they 
resist.  Then  it  is  readily  seen  that  the  stress  at  d,  represented  by  nd,  is 


and  at  e  it  is 


where  I  is  the  moment  of  inertia  of  the  cross-section  of  the  beam  at  the 
block  abed.    Then  the  average  increment  stress  between  d  and  e  is 

.„     A/  +  A/'      (S*x\(h  +  y\ 

f     ~ir  -\TP/vT/ 

Multiplying  this  by  bxde,  where  6  =  width  of  beam,  we  obtain  the  total 
horizontal  shear  along  ee,  which  is 


Now  this  shear  is  distributed  over  the  horizontal  strip  through  the  block 
at  ee.  If  b  be  the  width  of  the  beam,  the  area  of  this  horizontal  strip  will 
be  bbx.  Then  the  shear  per  square  inch  upon  it  will  be 


from  which  we  obtain  the  formula 


STRUCTURAL  ENGINEERING 


From  Formula  (G)  the  horizontal  shear  (and  incidentally  the  vertical) 
can  be  obtained  at  any  point  in  any  rectangular  beam  either  above  or 
below  the  neutral  axis. 

It  is  readily  seen  that  Formula  (G)  is  an  equation  to  a  parabola, 
which  means  that  the  horizontal  shear  on  any  rectangular  beam,  and 
likewise  the  vertical  shear,  varies  on  any  vertical  section  as  the  ordinates 
to  a  parabola,  as  represented  by  the  horizontal  ordinates  to  the  curve 
axb  shown  in  Fig.  73.  If  y  =  h,  the  shear  S^  is  equal  to  0.  If  #  =  0, 
S1  —  f  S/Af  which  is  the  maximum,  where  A  =  area  of  cross  section 
of  beam.  So  it  is  seen  that  both  the  vertical  and  horizontal  shear  is  0  at 
the  top  and  vlikewise  at  the  bottom  of  any  beam,  and  a  maximum  at  the 
neutral  axis.  The  same  is  true  of  all  beams. 

In  case  the  width  of  a  beam  varies  the  general  equation 


m 


8 


can  be  used  for  determining  the  horizontal  shear,  where  m  =  the  moment 
about  the  neutral  axis  of  the  part  of  the  cross  section  above  ee  or  below 
in  case  ee  is  below  the  neutral  axis  and  6  =  width  of  beam  at  ee. 

62.  Maximum  Stress. — It  is  seen  from  the  preceding  article  that 
every  particle  in  a  loaded  beam,  except  the  particles  at  the  top  and  bottom 
edges,  and  those  in  the  neutral  axis,  is  subj  ected  to  horizontal  and  vertical 
shear  and  to  a  direct  horizontal  stress  which  may  be 
either  tension  or  compression,  depending  upon  the 
position  of  the  particle  in  the  beam.  In  the  case  of 
a  cantilever  beam  the  particles  at  the  top  edge  are 
subjected  to  direct  tension  only,  while  all  particles 
between  the  top  edge  and  neutral  axis  are  subjected 
to  horizontal  and  vertical  shear  and  also  to  direct 
Fig.  74  tension,  and  the  particles  in  the  neutral  axis  are  sub- 

jected only  to  horizontal  and  vertical  shear.  At  the 
bottom  edge  of  a  cantilever  beam  the  particles  are 
subjected  to  a  direct  compression  only,  while  the 
particles  between  the  bottom  edge  and  neutral  axis 
are  subjected  to  horizontal  and  vertical  shear  and 
also  to  direct  compression.  In  the  case  of  a  simple 
beam  we  have  exactly  the  reverse  of  a  cantilever 
beam.  As  an  illustration,  let  AB  (Fig.  74)  represent 
a  short  portion  of  a  simple  beam  where  oo  represents 
the  neutral  axis.  The  particles  at  the  top  edge  are  subjected  to  direct 
compression  only,  as  indicated  at  a,  while  the  particles  between  the  top 
edge  and  neutral  axis  are  subjected  to  horizontal  and  vertical  shear  and 
also  to  direct  compression  as  indicated  at  6,  the  particles  in  the  neutral 
axis  are  subjected  only  to  horizontal  vertical  shear  as  indicated  at  c.  At 
the  bottom  edge  the  particles  are  subjected  to  direct  tension  only,  as  in- 
dicated at  e,  while  the  particles  between  the  bottom  edge  and  the  neutral 
axis  are  subjected  to  horizontal  and  vertical  shear  and  also  to  direct 
tension,  as  indicated  at  d. 

It  is  seen  from  Fig.  74  that  the  intensity  of  the  maximum  stress  on 
any  particle  at  the  top  or  bottom  edge  of  the  beam  is  simply  the  direct 
Stress  on  it,  but  the  intensity  of  the  maximum  stress  on  any  other  particle 


THEORETICAL  TREATMENT  OF  BEAMS  69 

is  not  so  evident,,  for  here  the  maximum  stress  is  due  to  the  combined 
action  of  the  shearing  forces  and  direct  stress.  Let  Fig.  75  represent  a 
small  imaginary  block  taken  from  a  simple  beam,  corresponding  to  the 
block  at  d,  Fig.  74.  For  convenience  consider  the  block  to  be  a  cube 
having  sides  of  unit  length,  and  let  h  represent  the  horizontal  shear  and 
v  the  vertical  shear.  It  is  readily  seen  that  the  h  along  fe  and  the  v  along 
ek  acting  against  the  v  along  fg  and  the  h  along  kg  produce  tension  upon 
all  such  strips  through  the  block  as  fk,  while  the  h  along  fe  and  the  v 
along  fg  acting  against  the  v  along  ek  and  the  h  along  kg  will  produce 
compression  upon  all  such  strips  through  the  block  as  eg.  Now  it  is 
evident  that  the  maximum  tension  and  compression  as  the  case  may  be, 
due  to  the  shearing  forces,  will  be  upon  strips  perpendicular  to  the 
resultant  of  the  shearing  forces.  As  the  horizontal  and  vertical  shear  are 
equal  it  is  obvious  that  their  resultant  will  always  be  at  45°  with  the 
horizontal  and  vertical  and  hence  the  maximum  tension  or  compression 
stresses  due  to  these  forces  will  be  upon  strips  perpendicular  to  this 
direction,  that  is,  45°  from  the  direction  of  the  shearing  forces  them- 
selves. It  is  readily  seen  that  the  direct  tensile  stress  indicated  as  p  on 
the  block  will  increase  the  tension  due  to  the  shearing  forces  on  the  strip 
fk  and  decrease  the  compression  on  the  strip  eg;  while  if  p  were  a  com- 
pressive  stress  just  the  reverse  would  be  true.  While  the  tensile  stress  p 
would  increase  the  tension  on  the  strip  fk,  yet  this  would  not  be  the 
maximum  tension  on  the  block,  for  evidently  the  maximum  would  be  on  a 
strip  through  the  block  perpendicular  to  the  resultant  of  the  stress  p  and 
the  shearing  forces  v  and  h.  This  resultant  would  have  some  position  as 
Rt,  if  p  were  tension,  and  Re,  if'p  were  compression,  and  the  maximum 
tensile  stress  then  would  be  upon  strips  through  the  block  perpendicular 
to  Rt;  and  if  p  were  compressive  the  maximum  compressive  stress  would 
be  upon  strips  through  the  block  perpendicular  to  Rc> 

If  the  resultant  of  the  shearing  forces  on  the  particles  of  any  beam 
were  graphically  combined  throughout  the  beam  with  the  direct  stresses 
on  the   particles,  the   balanced   resultants   obtained   would   form   curves 
known    as    the    lines    of    maximum 
stress.     As  an  example,  let  AB  (Fig.  L,  pg  P3 

76)    represent   a   simple  beam.      By   *  *    _         .• 

combining  the  forces  at  e,  d,  c,  etc., 

as   indicated   at   e' ',   d' ',   and    c',   the/\ 

curve  edc  would  be  obtained  and  by 

combining  the  forces  on  the  particles 

throughout  the  entire  beam  we  would 

obtain  some  such  lines  as  t,  tl,  t2, 

etc.,  representing  the  direction  of  the 

action     of     the     maximum     tensile 

stresses    in    the    beam.      Now,    evi-  Fig.  76 

dently,  if  any  particle  in  the  beam 

failed  in  tension  the  failure  would  take  place  perpendicular  to  these  lines 

of  maximum  tension. 

If  the  maximum  compressive  stresses  were  platted  in  the  same 
manner  they  would  take  some  such  position  as  indicated  by  the  dotted 
lines  on  the  beam,  and  if  any  particle  failed  in  compression  the  failure 


70  STRUCTURAL  ENGINEERING 

would    evidently   take   place   perpendicular   to   these   lines   of   maximum 
compression. 

It  is  readily  seen  from  Fig.  75  that  the  shear  due  to  the  shearing- 
forces  v  and  h  along  a  strip  as  eg,  45°  with  the  horizontal  or  vertical, 
will  be  zero,  for  the  two  shearing  forces  h  and  v  on  each  side  will  just 
balance  each  other  when  resolved  along  that  direction.  Then  the  only 
shear  along  that  direction  will  be  due  to  the  direct  stress  p.  If  we 
imagine  the  strip  rotated  about  o  it  is  readily  seen  that  the  shearing 
forces  will  produce  shear  along  the  strip  when  it  slopes  other  than  45° 
with  the  horizontal  and  the  maximum  shear  on  it  due  to  the  shearing 
forces  and  direct  stress  combined  will  occur  when  the  angle  of  slope  has 
some  particular  value.  But  this  angle  of  slope  depends  upon  the  relative 
intensity  of  the  shearing  forces  and  the  direct  stress.  The  direction  of 
maximum  tension  and  compression  stresses  also  depends  upon  the  relative 
intensity  of  the  shearing  forces  and  direct  stress.  So  we  will  now  deter- 
mine these  relations. 

Let  Fig.  77  represent  a  material  particle  assumed  rectangular, 
subjected  to  horizontal  and  vertical  shear  of  s  pounds  per  square  inch 
and  to  a  tensile  stress  of  /  pounds  per  square  inch.  Suppose  the  particle 
to  be  one  unit  in  thickness,  and  let  a  be  its  width 
and  b  its  length.  Then  sa  will  be  the  total  vertical 
shear  and  sb  the  total  horizontal  shear,  and  fa  the 
total  tensile  stress,  as  indicated  in  the  figure.  Let 
dd  be  an  imaginary  diagonal  strip  through  the  par- 
ticle. If  the  shearing  and  tensile  forces  be  resolved 
Fig.  77  perpendicularly  and  parallel  to  this  imaginary  strip 

dd,  the  components  perpendicular  to  the  strip  will 

produce  tension  on  it,  while  the  components  along  the  strip  will  produce 
shear.  Let  t  be  the  tensile  stress  per  square  inch  perpendicular  to  the 
strip  dd  and  let  v  be  the  shearing  stress  per  square  inch  along  the  strip, 
and  let  9  be  the  angle  that  the  strip  dd  makes  with  the  horizontal  axis 
of  the  particle;  then,  considering  the  forces  on  one  side  of  dd  only,  we 
have 

(1)  tdd  =  fa  sinO  +  sb  sinO  +  sa  cos0  for  the  tension, 
and 

(2)  vdd  =  fa  cosO  +  sb  cos6-sa  sin#  for  the  shear. 

Dividing  (1)  and  (2)  by  dd  and  substituting  sin0  for  a/dd  and  cos0  for 
b/ddt  we  have 


(3)  t  =  f  sin20  +  2s  cosOsinO  =  ~  (1  -  cos20)  +  *  sin20, 
and 

(4)  v  =  f  sinOcosO  +  s  (cos20  -  sin2^)  =  £•  sin2^  +  s  cos20. 


It  is  evident  that  t  will  be  a  maximum  when  6  has  a  certain  value  and  v  a 
maximum  when  6  has  a  certain  other  value.     Differentiating  (3)  we  have 

dt  =  f  sinZOdO  +  2s  cos26d6. 


THEOEETICAL  TREATMENT  OF  BEAMS  71 

Now,  t  will  be  a  maximum  when 

jg  =  f  sin20  +  2s  cos20  =  0, 
from  which  we  obtain 

(5)   tan  26  =  - j- 

as  the  value  of  0  when  t  is  a  maximum. 

Treating  (4)  in  the  same  manner  we  obtain 

(6)   tan  20  =  — 

as  the  value  of  0  when  v  is  a  maximum.     Expressing  the  value  of  the 
tangent  in  (5)  in  terms  of  the  sine,  we  have 

sin  20  _2s 

Vl-sin2  20~     T' 
Squaring  and  reducing,  we  have 

(a)   sin  20-  ±  —  9* 


Then  expressing  the  value  of  the  tangent  in  (5)  in  terms  of  the  cosine, 
we  have 

Vl-cos2  20        2s 


cos  26  f 

from  which  we  obtain 


(6)   cos20=± 


V/2  +  4s2 

Referring  to  equation  (5),  as  tan  20  =  -sin  20/cos  26,  a  minus  quantity,  it 
is  evident  that  sin  26  and  cos  26  will  have  opposite  signs,  that  is,  when 
one  is  plus  the  other  will  be  minus.  Then  substituting  the  value  of  sin  26 
and  cos  26  given  in  (a)  and  (6),  respectively,  in  (3),  taking  one  plus  and 
the  other  minus,  we  obtain 


By  treating  (6)  in  the  same  manner  as  (5)  was  treated  above,  we  obtain 

•n2  _/ 

"V/2 

and 

~  V/f  +  4*2 
Substituting  these  values,  which  will  have  like  sign,  in   (4),  we  obtain 


The  maximum  or  minimum  tensile  or  compressive  stresses  on  any 
material  particle  subjected  to  shear  and  direct  stress  can  be  computed 
from  (H)..  If  /  be  tension,  t  will  be  tension  when  the  radical  is  taken  as 


STRUCTURAL  ENGINEERING 


Fig.   78 


plus  and  compression  when  taken  as  minus.  Thus  both  tension  and 
compression  stresses  are  obtained.  If  /  be  compression,  by  using  the  plus 
sign  before  the  radical  we  obtain  the  compressive  stress  and  by  using  the 
minus  sign  we  obtain  the  tension.  As  the  plus  -sign  in  (H)  gives  the 
maximum  stress  in  all  cases  it  is  rarely  necessary  to  use  the  minus  sign. 
The  maximum  shear  on  any  material  particle  subjected  to  shear  and 
direct  stress  can  be  obtained  from  (I).  It  is  immaterial  whether  the 
plus  or  minus  sign  before  the  radical  be  used  as  the  two  values  will  be 
the  same. 

The  direction  of  the  maximum  shearing  stress  can  be  obtained  from 
(6)  and  a  direction  perpendicular  to  the  maximum  and  minimum  tensile 
and  compressive  stresses  can  be  obtained  from  (5). 

There  are  only  a  few  cases  where  the  above  formulas  need  to  be 
applied.  They  need  not  be  applied  in  the  case  of  ordinary  beams,  for 
here  both  the  maximum  tension  and  compression  stresses  occur  on  the 
outer  elements  where  the  shearing  stress  is  zero,  and  the  maximum  shear 
occurs  at  the  neutral  axis  where  the  direct  stress  is  zero. 

63.  Deflection  of  Beams. — Beams 
supporting  loads  deflect  or  bend  ow- 
ing to  the  distortion  of  their  parts  which 
results  from  the  loading.  Let  Fig.  78  rep- 
resent an  unloaded  rectangular  wooden 
cantilever  beam  which  we  will  assume  to 
be  straight  and  absolutely  horizontal. 
Imagine  this  beam  divided  into  very  short 
rectangular  blocks  by  imaginary  planes 
passing  perpendicularly  to  its  longitudinal 
axis.  Now  if  loads  were  applied  to  this 
beam,  these  rectangular  blocks  would  be- 
come wedge-shaped  as  shown  in  Fig.  79, 
and  the  imaginary  planes  dividing  the  beam 
into  blocks  instead  of  being  parallel  would 
become  tangent  to  some  curve  as  ss's",  and 
the  neutral  axis  oo  instead  of  being  straight 
would  become  a  curve  which  would  be 
known  as  the  "elastic  curve"  of  the  beam. 
The  distance  from  the  intersection  of  any 
two  adjacent  imaginary  planes  to  the  neu- 
tral axis  oo  would  be  known  as  the  radius 
of  curvature  of  the  elastic  curve  at  the  in- 
tercepted block.  The  vertical  distance  that 
any  point  on  the  neutral  axis  would  move 
down  from  its  original  position  would  be 
known  as  the  deflection  of  the  point  and  of 
the  beam  at  that  point,  and  the  deflection 
of  any  one  point  on  the  neutral  axis  in 
reference  to  another  would  be  the  differ- 
ence of  their  deflections.  For  example,  the 
distance  y  (Fig.  79)  from  the  horizontal 
line  Ho  to  the  point  g  would  be  the  deflec- 
Fig.  79  tion  of  point  g,  and  likewise  the  deflection 


THEORETICAL  TREATMENT  OF  BEAMS 


73 


of  the  beam  at  that  point,,  while  the  distance  z  would  be  the  deflection  of 
point  g  in  reference  to  point  k. 

It  is  readily  seen  that  the  curving  of  the  neutral  axis  results  from 
the  distortion  of  the  blocks.  Then  it  is  evident  that  if  the  slope  of  the 
blocks  in  reference  to  one  another  be  determined,  an  equation  of  the 
elastic  curve  expressing  this  relation  can  be  derived,  from  which  the 
deflection  of  the  beam  at  any  point  in  reference  to  any  other  point  can  be 
computed.  So  we  shall  now  proceed  to  derive  such  an  equation  of  the 
elastic  curve. 

Let  Fig.  80  represent  any  four  consecutive  distorted  blocks  of  the 
beam  shown  in  Fig.  79.  By  drawing  the  center  line  of  each  block  we 
have  the  broken  line  curve  BabcA  which  has  for  its  limit  the  elastic 
curve  SS  as  the  lengths  of  the  blocks  decrease;  that  is,  become  infini- 
tesimal. It  is  readily  seen  that  the  radius  of  curvature  at  any  block  is 
perpendicular  to  its  center  line 
at  midpoint.  So  the  slope  of  the 
center  line  of  any  block  with  the 
horizontal  or  X-axis  is  the  same 
as  that  of  the  tangent  of  the 
elastic  curve  at  that  point.  In 
fact,  the  two  coincide. 

Let  HB  be  a  horizontal 
line.  Then  </>  is  the  angle  that 
the  tangent  to  the  elastic  curve 
at  the  center  of  the  block  vurw 
makes  with  the  horizontal  or  X- 
axis.  Now,  as  the  length  of  the 
block  vurw  decreases,  the  center 
line  aB  comes  nearer  and  nearer 
to  coinciding  with  the  arc  of  the 
elastic  curve  passing  through  the 
block.  So  if  the  length  of  the 
block  be  infinitesimal  the  points 
a  and  B  will  be  on  the  elastic 
curve  and  the  vertical  distance 
ad  will  be  the  deflection  of  point 
a  in  reference  to  B,  and  ad 
would  be  a  first  differential  of  the  vertical  or  y  ordinate  to  the  elastic 
curve  at  that  point.  Likewise,  the  distance  eb,  gc,  and  JcA  will  each  be 
a  first  differential  of  a  y  ordinate  to  the  elastic  curve  when  the  length  of 
each  of  the  other  blocks  becomes  infinitesimal.  Letting  A  represent  the 
deflection  of  point  A  in  reference  to  B,  we  have 


Fig.   80 


which  is  simply  the  summation  of  the  first  differentials  summed  up  from 
B  to  A.  Let  dB  =  ea  =  gb  =  kc  =  dx,  an  infinitesimal,  which  would  really 
be  the  case  when  the  lengths  of  the  blocks  are  infinitesimal.  Then  by 
prolonging  aB,  the  center  line  of  block  vurw,  to  /  we  have  ef  =  ad.  Now 
as  ad  and  eb  are  each  a  first  differential  of  a  y  ordinate  to  the  elastic 
curve,  fb  is  a  second  differential  of  the  same,  being  the  difference  between 
two  consecutive  first  differentials;  and  similarly  he  and  mA  are  also 


74 


STRUCTURAL  ENGINEERING 


second  differentials  of  y  ordinates  to  the  elastic  curve.  It  is  readily  seen 
that  gc  =  ad  +  fb  +  hc  and  that  kA=ad  +  fb  +  hc  +  mA,  etc.;  that  is,  each 
first  differential  is  equal  to  the  summation  of  the  second  differentials  plus 
the  first  differential  ad  summed  up  from  B  to  the  point  considered.  Now 
if  the  summation  be  from  A  to  B  instead  of  B  to  A,  as  above,  we  have  the 
case  shown  in  Fig.  81,  where  ce,  bf,  ak,  and  Bn  are  first  differentials,  and 
gb,  ha,  and  mB  are  second  differentials.  Taking  A  as  the  origin,  we  have 
ak  =  ce-gb-ha,  and  Bn  =  ce-gb-ha-mB;  that  is,  any  first  differential 
of  a  y  ordinate  to  the  elastic  curve  is  equal  to  the  summation  of  the  second 
differentials  summed  up  from  the  origin  to  the  point  considered,  plus  the 
first  differential  ce.  (Fig.  81.)  The  second  differentials  are  minus  in  the 
last  case,  as  they  are  measured  down  to  the  curve;  while  the  first  dif- 
ferentials are  measured  up  to  the  curve.  For  the  deflection  of  point  A  in 

reference  to  B  we  have  A  =  ce  +  bf 
+  ak  +  Bn,  which  is  simply  the  sum- 
mation of  the  first  differentials 
summed  up  from  A  to  B.  Now,  as 
stated  above,  any  first  differential 
at  any  point  is  equal  to  the  summa- 
tion of  the  second  differentials 
summed  up  from  A  to  the  point, 
plus  the  first  differential  ce,  while 
in  the  case  shown  in  Fig.  80  the 
first  differential  at  any  point  is 
equal  to  the  summation  of  the  sec- 
ond differentials  summed  up  from  B 

Fig.  si  to  the  point,  plus  the  first  differen- 

tial ad.      It  is  readily  seen  that  as 

far  as  the  deflection  of  point  A  in  reference  to  point  B  is  concerned, 
both  of  the  first  differentials  ad  (Fig.  80)  and  ce  (Fig.  81)  are 
constants,  and  being  such  they  will  appear  as  constants  of  integration  in 
the  summation  of  the  second  differentials.  It  is  first  necessary  to  derive 
an  expression  for  the  second  differential  of  any  y  ordinate  to  the  elastic 
curve  in  terms  of  known  quantities.  Referring  to  Fig.  80,  if  the  length 
of  the  two  blocks  stuv  and  vurw  be  infinitesimal,  the  angle  subtended  by 
their  radii  of  curvature  will  be  an  infinitesimal  angle  d6,  and  as  one  radius 
is  perpendicular  to  aB,  the  center  line  of  block  vurw,  and  the  other  one 
to  ab,  the  center  line  of  block  stuv,  the  angle  fab  will  be  equal  to  that 
subtended  by  the  radii,  or  dO.  Then,  as  the  angle  dO  (Fig.  80)  is 
infinitesimal,  we  have 


=  abd6  or 


(1). 


The  permissible  deflection  of  a  beam  in  any  case  is  very  small  compared 
to  its  length  (1/1,000),  so  that  no  appreciable  error  will  be  made  if  we 
assume  the  slope  distance  ab  =  ea  =  dx.  Then  by  substituting  in  (1),  we 
have 


d2y  =  dxdO. 


(2). 

Now  dO  is  the  slope  that  the  block  stuv  makes  with  the  block  vurw, 
which  results  entirely  from  the  distortion  of  the  block  stuv.  Then, 
evidently,  the  value  of  dO  will  be  directly  proportional  to  the  bending 


THEOEETICAL  TREATMENT  OF  BEAMS  75 

moment  at  the  block  stuv  and  inversely  proportional  to  the  modulus  of 
elasticity  of  the  material  composing  the  block  and  also  inversely  propor- 
tional to  the  moment  of  inertia  of  its  cross-section.  Then  dO  can  be 
expressed  in  terms  of  these  quantities. 

Through  a  (Fig.  80)  draw  the  line  s't'  parallel  to  st.  Then  the 
angle  s'a^-dQ.  Let  a  =  the  longitudinal  distortion  of  an  element  of  the 
block  (stuv)  distance  2  above  ab.  Then  we  have 


But,  according  to  Art.  33, 

Mz     ab      Mzdx 


(Mz/I  —  stress  on  element)  (see  Formulas  B  and  D,  Arts.  33  and  53), 
where  M  =  the  bending  moment  of  the  block  stuv  and  /  =  moment  of 
inertia  of  the  cross-section  of  the  block  and  E  the  modulus  of  elasticity 
of  the  material  composing  it.  Now,  combining  these  two  equations,  we 
have 

Mdx 

Substituting  this  value  of  dO  in  (2)  we  have 

AMH 

-IE" 

which  is  the  expression  for  the  second  differential  of  the  y  ordinate  to  the 
elastic  curve  at  block  stuv  in  terms  of  known  quantities,  but  which  at  the 
same  time  is  really  the  expression  for  the  vertical  drop  of  the  elastic 
curve  from  a  to  b  due  wholly  to  the  distortion  of  block  stuv.  Now,  this 
same  expression  will  hold  in  the  case  of  any  of  the  blocks,  as  no  special 
case  was  taken.  Then,  evidently,  by  substituting  the  proper  value  of  M 
in  the  expression  and  integrating  twice,  the  deflection  of  point  A  in 
reference  to  B  will  be  obtained.  It  is  evident  that  the  above  discussion 
will  apply  to  any  number  of  blocks  as  well  as  to  four,  so  our  expression 
for  the  second  differential  of  the  y  ordinates  to  the  elastic  curve  will  apply 
to  the  entire  beam  or  any  part  of  it,  as  A  and  B  can  be  any  two  points, 
and  as  the  bending  in  the  case  of  any  beam  can  be  nothing  different  from 
the  bending  of  a  cantilever  beam  here  considered,  the  expression  is 
evidently  applicable  in  the  case  of  any  beam  whatever,  loaded  in  any 
manner,  and  is  hence  a  general  differential  equation  of  the  elastic  curve 
of  any  beam  or  any  body  whatever,  subjected  to  bending  stresses.  For 
convenience,  the  equation  is  usually  written 


Referring  to  Fig.  79,  it  will  be  seen  that  the  curve  ss's"  is  an 
evolute,  while  the  elastic  curve  oo  is  an  involute.  In  case  of  a  simple 
beam  there  would  be  two  branches  to  the  evolute. 

64.    Deflection  of  Cantilever  Beams. — 

Case  I.     When  the  beam  supports  a  single  load  at  its  free  end. 


76 


STRUCTURAL  ENGINEERING 


Let  AB  (Fig.  82)  represent  the  beam  of  length  L,  P  the  load,  and 
let  x  and  y  be  the  co-ordinates  to  the  elastic  curve  at  any  point  when  the 
origin  is  taken  at  B,  and  x,  and  y,  the  co-ordi- 
nates when  the  origin  is  taken  at  A. 

First,  taking  B  as  the  origin,  we  have  M—P 
(L-x)  for  the  bending  moment  at  any  point  x 
distance  from  B.  Now  substituting  this  value  of 
M  in  Formula  (K)  (Art.  63),  we  have 


Fig.   82 


Integrating  once,  we  have 


dx 


2 


Now,  dy,/dx,  is  the  expression  for  the  tangent  of  the  angle  that  the  elastic 
curve  makes  with  the  horizontal  or  x  axis  at  any  point  x  distance  from  B. 
It  is  readily  seen  that  dy/dx  =  0  at  B,  the  origin,  as  the  elastic  curve  is 
horizontal  at  that  point.  But  x  —  0  also  at  that  point.  Then  substituting 
0  for  dy/dx  and  for  x  in  the  above  equation,  we  have  C^O.  Now  as 
Ci  =  0,  the  preceding  equation  becomes 


which  is  the  general  equation  for  the  slope  of  the  elastic  curve  at  any 
point  between  B  and  A  with  B  as  the  origin.  From  this  equation,  the 
slope  of  the  beam  ( =  slope  of  the  elastic  curve)  at  any  point  can  be 
computed  by  substituting  for  x  its  numerical  value.  For  example,  the 
slope  of  the  beam  at  a  point  b  distance  from  B  is 


dy 
dx 


El 


which  is  the  tangent  of  the  angle  that  the  tangent  to  the  elastic  curve  at 
that  point  makes  with  the  .r-axis. 
Next  integrating  (2)  we  have 


Now  at  B  both  y  and  x  =  0.     Then  substituting  0  for  x  and  also  for  y,  we 
have  C2  =  0.     Hence,  we  have 


which  is  the  algebraic  equation  of  the  elastic  curve  when  the  origin  is  at 
B.  From  this  equation  the  deflection  of  the  beam  at  any  point  can  be 
computed  by  substituting  for  x  its  numerical  value.  It  is  readily  seen 
that  the  deflection  will  be  a  maximum  when  x  =  L.  Let  A  =  the  maximum 
deflection;  then  we  have 


THEORETICAL  TREATMENT  OF  BEAMS  77 

PL3 
3El' 

Now  taking  A  as  the  origin,  we  have  for  the  bending  moment  at  any 
point  x,  distance  from  A,  M  =  Px,.  Then  substituting  this  value  of  M  in 
Formula  (K)  (63),  we  have 


Integrating  once,  we  have 


Now,  dy,/dx,  —  0  when  x  =  L.  Then  substituting  these  values  for  dy,/dxf 
and  ,r,  respectively,  in  the  preceding  equation,  we  have  C'  =  -PL2/2. 
Then  substituting  this  value  of  C"  in  the  equation,  we  have 


Px. 


dx. 


for  the  general  equation  for  the  slope  of  the  elastic  curve,  from  which  the 
slope  of  the  curve  at  any  point  x,  distance  from  A  can  be  computed  by 
substituting  for  x,  its  numerical  value. 
Next,  integrating  (4),  we  have 

P™  3      pr  2 „ 

rx,      fL,  x,      ~ff 


6  2 

Now,  when  x,  -  0,  y,  -  0.     Substituting  0  for  x,  and  also  for  y,  we  have 
c"  =  0.     Then  we  have 


_  P  /x?      L2x\ 
~£/\6        >.  / 


(5) 


for  the  equation  of  the  elastic  curve  when  the  origin  is  at  A,  from  which 
the  vertical  ordinate  y,  at  any  point  x,  distance  from  A  can  be  computed 
by  substituting  for  x,  its  numerical  value.  Th'en,  by  subtracting  this  y, 
ordinate  from  the  maximum  deflection  A  the  deflection  of  the  point  in 
question  is  obtained.  It  is  readily  seen  from  Fig.  82  that  the  y,  ordinate 
is  equal  to  A  when  x,  -  L.  So  substituting  in  (5)  we  have 

P  /L3     L3\        PL3 


to  the 


3EI 

which  is  the  same  as  found  above  except  for  the  sign,  which  is  due 
ordinates  being  measured  in  the  opposite  direction. 

Case  II.     When  the  beam  supports  two  loads. 

Let  AB  (Fig.  83)  represent  the  beam,  P 
and  P'  the  loads,  and  let  a  be  the  distance  that 
load  P  is  from  B,  b  the  distance  between  the 
loads  and  d  the  distance  that  the  load  P'  is  from 
A,  the  end  of  the  beam.  Take  B  as  the  origin, 
and  let  x  and  y  represent  the  co-ordinates  to 
the  elastic  curve.  When  x  is  less  than  a,  we 
have  for  the  bending  moment  at  a  point  x  dis-  Fig.  83 


78  STRUCTURAL  ENGINEERING 

tance  from  B 

M=(a-x}P+(b  +  a-x}P', 

therefore,  EI^^  (a- x}P  +  (b  + a- x}P'. 

Integrating,  we  have 


. 

2  2 

But  dy/dx  =  0  when  x  -  0  ;  therefore,  C"  =  0,  and  we  have 


2 
Now  integrating  (6),  we  have 

r    Pa,-3     P'bx2      P'ax2      P' 


(6). 


But  y  =  0  when  x  =  0 ;  therefore,  C"  =  0,  and  we  have 


y   r  7 


*'ax2     P'x'\ 
2  6   / 


6  2 

Now  when  x  is  greater  than  a  and  less  than  a  +  b,  we  have 


for  the  bending  moment  at  any  point  between  P  and  P',  .r  distance  from 
B.    Therefore, 

EId^L=P 

dx2 

Integrating,  we  have 

dii  P'x2 

EI-±=P'ax  +  P'bx  -       -+  C,  ........................  (8). 

6/iT  2 

Now  it  is  readily  seen  that  equations  (6)  and  (8)  are  equal  when 
x  —  a  in  each.  So,  substituting  a  for  x  in  each,  equating  and  cancelling, 
we  have 


Substituting  this  value  of  C,  in  (8)  we  have 


. 

Integrating,  we  have 

P'ax*     P'bx2     P'x*     PaT-x 


Now  it  is  readily  seen  that  equations  (7)  and  (10)  are  likewise 
equal  when  x  =  a  in  each.  So  substituting  a  for  x  in  each,  equating  and 
cancelling,  we  have 

Pa* 
C"~     ~6~* 


THEORETICAL  TREATMENT  OF  BEAMS 


79 


Substituting  this  value  of  C,,  in  (10),  we  have 


1   IP 


?2     P'bx2 

-  +  — « — 


P'*3     Pa*x 


Now  the  slope  of  the  elastic  curve  at  any  point  between  B  and  the 
load  P  can  be  computed  from  equation  (6)  and  at  any  point  between  the 
loads  from  (9),  while  the  deflection  at  any  point  between  B  and  the  load 
P  can  be  computed  from  equation  (7)  and  at  any  point  between  the  loads 
from  (11)  by  substituting  for  x  its  numerical  value  in  each  case. 

The  maximum  deflection  will  be  at  A.  This  can  be  determined  in 
the  following  way: 

First  compute  the  deflection  of  the  beam  at  the  load  P'  from  equa- 
tion (11)  (substituting  (a  +  fe)  for  x).  Then  compute  the  slope,  that  is, 
the  tangent  of  the  slope  angle  at  that  point  from  (9)  and  multiply  this 
slope  by  the  distance  d,  which  will  give  the  deflection  of  the  point  A  in 
reference  to  the  point  at  P'.  Then  add  this  deflection  to  the  deflection  of 
the  beam  at  P'  ',  and  we  will  have  the  total  deflection  at  A. 

Case  III.  When  the  beam  supports  a  uniform 
load. 

Let  AB  (Fig.  84)  represent  the  beam  support- 
ing a  uniform  load  of  w  pounds  per  foot  of  length. 
Take  B  as  the  origin,  and  let  x  and  y  represent  the 
co-ordinates  to  the  elastic  curve.  Then  the  bending 
moment  at  any  point  x  distance  from  B  is 


Pigr.  84 


therefore, 


Integrating  once,  we  have 


When  x  =  0,  dy/dx  =  0  ;  therefore,  C'  =  0,  and  we  have 


Integrating  again,  we  have 


Now,  when  a?  =  0,  0  =  0;  therefore,  C"  =  0,  and  we  have 


. 

y~ 


2EI       2 


(13). 


From  equation  (12)  the  slope  at  any  point  of  the  beam  can  be 
computed,  and  from  (13)  the  deflection  at  any  point  can  be  computed  by 
substituting  for  x  its  numerical  value  in  each  case.  It  is  evident  that  the 
maximum  deflection  will  be  at  A,  the  free  end.  For  that  point  x  =  L. 


30  STRUCTURAL  ENGINEERING 

Then  substituting  L  for  x  in  equation  (13),  we  have 

wL* 

y~     ~ 


for  the  maximum  deflection. 

65.    Deflection  of  Simple  Beams. — 

Case  1.      •Vhen  the  beam  supports  a  single  load  at  mid-span. 

L  Let  AB    (Fig.   85)    represent   a   simple  beam  of 

^  ~~*|       length  L  supporting  a  single  load  P  at  mid-span. 

^J^^_.np    ^-~---\3    Let  R  and  R'  represent  the  reactions  at  A  and  B, 
^    ^ — ^^ \K     respectively,  which  are  equal  in  this  case.     Take  A 
Fig.  85  as  the  origin,  and  let  x  and  y  represent  the  co-ordi- 

nates to  the  elasti/1  curve.     For  the  bending  moment 
at  any  point  to  the  left  of  the  load,  x  distance  from  A,  we  have 


therefore,  we  have 
Integrating  once,  we  have 


dx        4 

Now  dy/dx  =  0  when  x  —  L/2.     Substituting  this  value  for  x,  we  have 


16 

and  substituting  this  value  of  C"  in  the  above  equation,  we  have 
EIdy_Px*     PL- 

>LTx-^r  TG~- 

Integrating  this  equation,  we  have 

Px'      P^x 

~-       " 


Now,  y  =  0  when  x  —  0  ;  therefore,  C"  -  0,  and  we  have 


which  is  the  equation  of  the  elastic  curve  to  the  left  of  the  load.  As  the 
elastic  curve  is  symmetrical  about  the  load,  there  is  no  need  for  deriving 
the  equation  for  the  curve  to  the  right  of  the  load.  However,  this  can  be 
readily  accomplished  by  substituting  in  Formula  (K)  (Art.  63),  the 
expression  for  the  bending  moment  to  the  right  of  the  load,  which  is 


It  is  readily  seen  that  the  maximum  deflection  will  occur  under  the 
load.  So  letting  A  represent  the  maximum  deflection,  and  substituting 
L/2  for  x  in  equation  (2),  we  have 


THEORETICAL  TREATMENT  OF  BEAMS  81 

PL3 


The  slope  of  the  elastic  curve  at  any  point  between  A  and  the  load  can  be 
computed  from  equation  (1). 

Case  II.      When  the  beam  supports  a  single  load  at  any  point. 

Let  AB  (Fig.  86)  represent  a  simple  beam  7 

supporting  a  load  P  at  any  point  z  distance 
from  A.  Let  R  and  R'  represent  the  reactions 
at  A  and  B,  respectively,  due  to  this  load  P. 
Take  A  as  the  origin,  and  let  x  and  y  represent  p. 

the  co-ordinates  to  the  elastic  curve.  For  the 
bending  moment  at  any  point  to  the  left  of  the  load,  we  have 

j 

~~j 
therefore,  we  have 

Integrating,  we  have 


Integrating  again,  we  have 

Px*     Pzx3 


Now,  y  =  0  when  x  -  0  ;  therefore,  C"  -  0,  and  we  have 


Now  for  the  bending  moment  at  any  point  to  the  right  of  the  load,  that  is, 
when  #>£,  we  have 


+Pz, 
i 

therefore,  we  have 


Integrating,  we  have 


Integrating  again,  we  have 

,,  +  C2  ........................  ..(6). 


It  is  readily  seen  that  equations  (3)  and  (5)  are  equal  when  x  —  z 
in  each.  Then  substituting  z  for  x  in  each  equation,  equating  and 
cancelling,  we  have 


STRUCTURAL  ENGINEERING 


Now  substituting  this  value  of  Cx  in  (6),  we  have 

Pvr-       Prrs  p^-r 

'  ^-'-^  ....................  w- 


It  is  readily  seen  that  equations  (4)  and  (8)  are  equal  when  x  —  z 
in  each  case.  Then  substituting  z  for  x  in  each,  equating  and  cancelling, 
we  have 


Now  it  is  readily  seen  that  y  in  equation  (8)  equals  zero  when 
x  —  L.  Then  substituting  L  for  x  and  Pz3/6  for  C2  in  that  equation, 
equating  and  reducing,  we  have 

Pz°-     PzL          > 

; 


Substituting    this    value    of    C"    in  equations    (3)    and    (4),    we    have, 
respectively, 

dy  __Px*     Pzx*     Pz*  PzL 

^'-  ~~  ~  ~~  ~~ 


Pz-x     PzxL     Pzzx 


Now,  from  equation  (11)  the  slope  of  the  elastic  curve  at  any 
point  to  the  left  of  the  load  can  be  computed,  while  the  deflection  at  any 
point  to  the  left  of  the  load  can  be  computed  from  (12). 

From  equations  (7)  and  (10)  we  have 

PzL      Pz* 

Cl~       3     "  6L' 
and  from  (9)  we  have 

c  -p*3- 

C2~~6~ 
Substituting  these  values  in  equations  (/5)  and  (G),  we  have,  respectively, 


EId-S.-P-,     —     ^     —  (13) 

dx~          ~  2L          3     ~  6L  ' 

Pzx*     Pzx3     PzxL     Ps*x     Pz* 


Now,  from  equation  (13)  the  slope  of  the  elastic  curve  at  any  point  to 
the  right  of  the  load  can  be  computed,  while  the  deflection  at  any  point  to 
the  right  of  the  load  can  be  computed  from  (14). 

Equations  (11)  and  (12)  apply  only  to  the  part  of  the  elastic  curve 
to  the  left  of  the  load,  while  equations  (13)  and  (14)  apply  only  to  the 


THEORETICAL  TREATMENT  OF  BEAMS  83 

part  to  the  right  of  the  load,  an<jl  hence  we  can  consider  the  elastic  curve 
to  be  composed  of  two  separate  curves  which  are  tangent  at  the  load. 
Now,  if  the  beam  supported  two  loads  some  distance  apart,  the  elastic 
curve  would  be  composed  of  three  separate  curves;  and  if  it  supported 
three  loads,  the  elastic  curve  would  be  composed  of  four  separate  curves; 
and  so  on.  That  is  to  say,  if  there  be  n  loads,  there  will  be  (w  +  1) 
separate  curves  composing  the  elastic  curve.  The  equations  for  each  of 
these  curves  could  be  derived  as  readily  as  the  above  equations,  and  then 
from  these  equations  the  slopes  and  deflections  of  the  elastic  curve  could 
be  computed.  The  main  labor  is  the  determining  of  the  constants  of 
integration.  It  will  be  observed  that,  at  the  point  of  maximum  deflection, 
dy/dx  =  0,  as  the  tangent  to  the  curve  will  be  horizontal  at  that  point. 

Case  III.  When  the  beam  supports  a  uni- 
form load. 

Let  AB  (Fig.  87)  represent  a  simple  beam, 
length  L,  supporting  a  uniform  load  of  w  pounds 
per  foot  of  length.  Let  R  and  Rl  represent  the  Fi&  87 

reactions  at  A  and  B,  respectively,  and  let  x  and 
y  represent  the  co-ordinates  of  the  elastic  curve,  A  being  taken  as  the 
origin.     Then  the  bending  moment  at  any  point  x  distance  from  A   is 


therefore, 

d'2y  __ 

(*to?s 

Integrating  once,  we  have 


_  . 

dx         4  .          6 

Now,  dy/dx  -  0  when  x  -  L/2,  as  the  tangent  to  the  elastic  curve  is 
horizontal  at  that  point.     Now  substituting  this  value  of  x,  we  have 


and  substituting  this  value  of  C'  in  the  above  equation,  we  have 

dy  _wLx2      wx5     wL3 
'1~~~    '~~~~' 


Integrating  this  equation,  we  have 

wLx3      wx*     wL3x 


But  y  -  0  when  x  -  0  ;  therefore,  C"  =  0,  and  we  have 


STRUCTURAL  ENGINEERING 


y    EI\    13         24  "     24  /; 


,r.  0»  ->,,., 
(2), 


from  which  the  deflection  at  any  point  can  be  computed. 

It  is  evident  that  the  deflection  will  be  a  maximum  at  midspan  ;  that 
is,  when  x  —  L/2.  So  letting  A  represent  the  maximum  deflection,  and 
substituting  L/2  for  a;  in  (2),  we  have 


66.  Proposition.  —  The  bending  moment  at  any  point  in  a  beam  is 
equal  to  the  bending  moment  at  any  other  point  plus  the  shear  at  the 
other  point  multiplied  by  the  distance  between  the  points,  plus  the 
algebraic  sum  of  the  moments  of  the  forces  between  the  points  about  the 
point  in  question. 

Let  AB  (Fig.  88)  represent  a  simple  beam  supporting  the  loads 
P,  PI,  and  P2  as  shown.  Now,  according  to  the  above  proposition,  the 


t 


Fig.   89 

bending  moment  at  D  is  equal  to  the  bending  moment  at  E,  plus  the  shear 
at  E  multiplied  by  c,  plus  the  sum  of  the  moments  of  the  forces  between 
the  points  about  D.  This  is  readily  seen  to  be  true,  for,  taking  moments 
about  E,  and  letting  R  represent  the  reaction  at  A  due  to  the  three  loads, 
we  have 


for  the  bending  moment  at  that  point,  and  taking  moments  about  D,  we 
have 

dPl-  hP2 

-dPl-hP2, 

which  is  seen  to  agree  with  the  above  proposition,  as  the  part  R(a  +  b)  - 
Pb  is  the  bending  moment  at  E,  R—P  is  the  shear  at  E,  which  is  multi- 
plied by  c,  the  distance  between  E  and  D,  and  -dPl  -  hP2  is  the  sum  of 
the  moments  of  the  forces  between  E  and  D,  about  D.  The  minus  signs 
in  the  last  are  simply  the  signs  of  the  moments. 

The  above  proposition  is  readily  proven  in  the  case  of  any  beam 
whatever,  but  regardless  of  its  simplicity,  it  is  quite  useful. 

67.  Reactions,  Shears,  and  Bending  Moments  on  Overhanging 
Beams.  —  Let  ABC  (Fig.  89)  represent  an  overhanging  beam  supported 
at  B  and  C  and  which  in  turn  supports  the  loads  P  and  P',  as  shown. 
The  part  AB  is  known  as  the  cantilever  arm,  while  the  part  BC  is  known 
as  the  anchor  arm.  Let  R  represent  the  reaction  at  B  due  to  the  two 
loads,  P  and  P',  and  let  R'  represent  the  reaction  at  C  due  to  the  same. 
Now,  taking  moments  about  C,  we  have 


THEORETICAL  TREATMENT  OF  BEAMS 


from  which  we  obtain 

aP  +  bP' 
~~ 


Then  taking  moments  about  B,  we  have 

R' 
from  which  we  obtain 


If  eP  be  greater  than  dP'  it  is  evident  that  R'  will  act  downward 
upon  the  beam,  and  hence  the  beam  would  pull  upward  upon  the  support 
at  C,  which  would  require  that  the  beam  be  anchored  in  some  manner  to 
the  support.  In  all  such  cases  the  reaction  is  spoken  of  as  being  negative. 
If  dP'  be  greater  than  eP,  of  course  R'  will  be  positive,  the  same  as  for 
simple  beams. 

In  case  of  a  greater  number  of  loads,  the  reactions  are  determined 
in  the  same  manner  as  shown  above  for  the  two  loads.  We  simply  include 
the  moment  of  each  load  in  the  equation  of  moments.  In  the  case  of  a 
uniform  load  the  reactions  are  determined  practically  in  the  same  manner, 
except  that  we  use  average  lever  arms  for  the  uniform  load.  For  example, 
suppose  the  beam  ABC  (Fig.  89)  supports  a  uniform  load  of  w  pounds 
per  foot  of  length,  extending  from  A  to  C.  Let  R,  represent  the  reaction 
at  B  due  to  this  uniform  load,  and  let  R,,  represent  the  reaction  at  C  due 
to  the  same.  Then  taking  moments  about  C,  we  have 


and  taking  moments  about  B,  we  have 

'>v...,.'   ±JB"=i(V-V)'\, .;;...;'";/"•' . 

In  case  the  beam  overhangs  two  supports  the  reactions  are  deter- 
mined as  above  by  taking  moments  about  the  supports  and  simply 
including  the  moment  of  each  force  acting  upon 
the  beam  in  the  equation  of  moments.  As  an  a  ^  b  d 

example,  let  ABCD  (Fig.  90)  represent  a  beam  ±9^f>. 

overhanging  two  supports.     Let  P,  Pf ,  and  P"  ^c 

represent  three  loads  supported  by  the  beam  as 

shown,  and  let  R  represent  the  reaction  at  B  Fig  90 

due  to  these  three  loads,  and  let  R'  represent 

the  reaction  at  C  due  to  the  same.    Now,  taking  moments  about  B,  we  have 

aP  -  bP'  ±  R'L  -  (e  +  L)P"  =  0, 
from  which  we  obtain 

±R'  = 


86  STRUCTURAL  ENGINEERING 

Then  taking  moments  about  C,  we  have 

-(L  +  a)P±RL-dP'  +  eP"  =  Q 
from  which  we  obtain 

(L  +  a)P  +  dP'-eP" 

- 


Now  suppose  the  beam  represented  in  Fig.  90  supports  a  uniform 
load  of  w  pounds  per  foot  of  length  extending  from  A  to  -D,  and  let  R, 
represent  the  reaction  at  B  due  to  this  uniform  load  and  let  R,,  represent 
the  reaction  at  C  due  to  the  same.  Then  taking  moments  about  B,  we 
have 


_ 


2L 

and  taking  moments  about  C,  we  have 

_ 


2L  2L 

The  determining  of  the  shear  at  any  section  of  an  overhanging 
beam  is  just  the  same  as  for  any  other  beam;  that  is,  the  shear  at  any 
section  is  equal  to  the  algebraic  summation  of  the  forces  on  either  side  of 
the  section  summed  up  from  the  end  of  the  beam  in  each  case.  For 
example,  the  shear  at  section  ss  of  the  beam  ABCD  (Fig.  90)  due  to  the 
three  loads  P,  P'  ,  and  P",  is 

S  =  P±RorP"±R'  +  P'. 

The  determining  of  the  bending  moment  at  any  section  of  an  over- 
hanging beam  is  just  the  same  as  for  any  other  beam;  that  is,  the  bending 
moment  at  any  section  is  equal  to  the  algebraic  summation  of  the  moments 
of  the  forces  on  either  side  of  the  section  about  the  section.  For  example, 
the  bending  moment  about  the  section  ss  of  the  beam  ABCD  (Fig.  90) 
due  to  the  three  loads  P,  P',  and  P"  is 


All  of  the  above  holds  as  well  for  cantilever  arms  as  for  anchor  arms, 
but  it  is  more  convenient  to  treat  the  cantilever  arms  as  independent 
cantilever  beams. 

68.    Reactions,  Shears,  and  Bending  Moments  on  Beams  Fixed 
at  One  End  and  Supported  at  the  Other.— Let  AB  (Fig.  91)  repre- 
sent such  a  beam  which  has  the  simple  support  at  A  and  the  fixed  support 
at  B.    As  the  beam  is  fixed  at  B  it  is  evident  that  any 
load  on  the  beam  will  produce  tension  in  the  elements 
above   the   neutral   axis   and   compression   below   the 
neutral  axis  at  that  point.     But  it  is  equally  evident 
that  the  reverse  is  true  just  to  the  right  of  the  simple 
Fig.  91  support  at  A;  that  is,  the  top  elements  of  the  beam  just 

to  the  right  of  A  will  be  in  compression  and  the  bot- 
tom ones  in  tension.  Then,  evidently,  there  will  be  some  section  be- 
tween A  and  B  where  the  bending  reverses;  that  is,  a  section  where 
there  is  no  bending,  and  hence  the  bending  moment  at  that  point  of  the 


THEOEETICAL  TEEATMENT  OF  BEAMS  87 

beam  will  be  equal  to  zero.  Let  S  be  such  a  point.  Then  the  part  SB  of 
the  beam,,  to  the  right  of  S,  would  be  simply  a  cantilever  beam,  while  the 
part  SA  of  the  beam.,  to  the  left  of  S,  would  be  a  simple  beam,  whence  the 
bending  or  curving  of  the  beam  due  to  loads  would  be  as  indicated  in 
Fig.  91.  The  point  S,  where  the  bending  moment  equals  zero,  would  be 
known  as  the  point  of  "contra-flexure,"  also  as  the  "point  of  inflection." 
As  a  case  of  concentrated  loads,  let  R  (Fig.  91)  represent  the 
reaction  of  A  due  to  a  single  load  P  at  any  distance  kL  from  A,  k  being 
any  fraction  less  than  unity.  Owing  to  the  unknown  forces  acting  upon 
the  beam  to  the  right  of  B,  this  reaction  R,  at  A,  cannot  be  determined  in 
the  usual  way  by  taking  moments  about  B.  Take  A  as  the  origin,  and  let 
x  and  y  represent  the  co-ordinates  to  the  elastic  curve.  Then,  for  the 
bending  moment  at  any  point  to  the  left  of  the  load,  that  is,  when  x  is 
less  than  kL,  we  have 

M  =  Rx, 

therefore, 


dx* 
Integrating  once,  we  have 


Integrating  again,  we  have 


But,  y  =  0  when  x  =  0  ;  therefore,  C"  =  0,  and  we  have 

C'x  ....................................  (2). 


Now  for  the  bending  moment  at  any  point  to  the  right  of  the  load, 
that  is,  where  x  is  greater  than  kL,  we  have 

M^Rx 

therefore, 

El  ^=Rx- 
dx- 

Integrating  once,  we  have 

du     Rx2     Px2 


But,  dy/dx  =  0  when  x  -  L.     Then  substituting  L  for  x,  and  equating, 
we  have 

RL>     PL* 
._  +_-«/,  -Cr 

Substituting  this  value  of  Cl  in  the  last  equation,  we  have 


88  STRUCTURAL  ENGINEERING 

Then  integrating  (3),  we  have 

Rx3     Px3     PkLx2      RL2x      PL2x 


2 

But  here  y  =  0  when  x  -  L.     Then  substituting  L  for  x,  equating  and 
cancelling,  we  have 

PL3 


3          3  2 

Substituting  this  value  of  C2  in  the  last  equation,  we  have 
Rx3^     Px^     PkLx2      RL2x      PL2x 

RL3     PL3     PkL3 


It  is  readily  seen  that  equations  (1)  and  (3)  are  equal  when  x  =  kL 
in  each.  Then  substituting  kL  for  x  in  each  equation,  equating  and 
reducing,  we  have 


Then  substituting  this  value  of  C"  in  (2)  we  have 
Rx3     Pk2L2x     RL2x     PL2x 


It  is  readily  seen,  also,  that  equations  (4)  and  (5)  are  equal  when 
x  —  kL  in  each.  Then  substituting  kL  for  x  in  each  of  these  equations, 
equating  and  reducing,  we  have 

B=f(*»-3*  +  2)  ...................................  (6), 

r  i  *  *  ^  r> 

from  which  the  reaction  at  A  due  to  a  load  at  any  point  on  the  beam  can 
be  computed.  For  example,  suppose  the  load  P  is  at  mid-span.  Then 
k  =  ^.  Substituting  this  ^  in  equation  (6)  we  have 


Again,  suppose  the  load  is  at  the  quarter  point  nearest  A.     Then 
k  =  £.    Substituting  this  J  for  A:  in  (6)  we  have 


If  there  be  more  than  one  load  on  the  beam,  the  reaction  at  A  due  to 
each  load  would  be  computed  separately  and  all  added  together  and  we 
would  thus  obtain  the  reaction  at  A  due  to  all  the  loads.  Of  course  the  k 
for  each  would  be  different  from  the  others;  that  is,  no  two  A;'s  would  be 
the  same. 

In  case  the  beam  supports  a  uniform  load  of  w  pounds  per  foot  of 
length  extending  over  the  full  length  of  the  beam,  that  is,  from  A  to  B, 


THEOEETICAL  TREATMENT  OF  BEAMS 
we  have 


_  . 

for  the  moment  at  any  point  x  distance  from  A,  where  R  represents  the 
reaction  at  A  due  to  this  uniform  load.     Therefore,  we  have 

£/*f=R*_*£!. 

dx2  2 

Integrating  once,  we  have 


. 

dx        2          6 

But  dy/dx  —  0  when  x  =  L.     So  substituting  L  for  x  in  the  preceding 
equation,  we  have 

RL*     wL° 

1~  +  ~T' 

Then  substituting  this  value  of  C"  in  the  last  equation,  we  have 


dy_R^     «£»     RL* 

>L  dx~    2      '    6          2  6 

Integrating  this  equation,  we  have 


But  here  y  —  0  when  x  —  0  ;  therefore,  C"  —  0,  and  we  have 


Now  y  —  0  in  (8)  when  x  -  L.     Then  substituting  L  for  x  (8),  we  have 

flL3      wL4     flL3     wL4 

~6~  ~~W  '    "2"     "6~J 
from  which  we  obtain 


That  is,  when  a  beam  supported  at  one  end  and  fixed  at  the  other 
supports  a  uniform  load,  the  reaction  at  the  supported  end  is  three- 
eighths  of  the  total  load. 

After  the  reaction  at  the  supported  end  due  to  the  loads  supported 
is  determined,  the  shear  and  bending  moment  at  any  point  in  the  beam 
are  obtained  just  the  same  as  in  the  case  of  a  simple  beam,  except  we  deal 
only  with  the  supported  end.  The  shear  at  any  point  is  equal  to  the 
reaction  at  the  supported  end  minus  all  intervening  loads,  while  the 
bending  moment  at  any  point  is  equal  to  the  reaction  at  the  supported 
end  multiplied  by  its  lever  arm,  minus  the  moment  of  all  intervening 
loads  about  the  point.  For  example,  the  shear  at  any  point  x  distance 
from  the  supported  end  due  to  a  uniform  load  of  w  pounds  per  foot  of 
span  is 

g 

S-  —  wL-wx, 
8 


90  STRUCTURAL  ENGINEERING 

while  the  bending  moment  is 


3               tvx 
M  —  —wLx — 

o  & 

If  x  -  Li,  we  have 


which  is  the  bending  moment  at  the  support  B  or  the  fixed  end. 

Deflections  can  be  computed  from  equations  (4),  (5),  and  (8). 
69.    Shears  and  Bending  Moments  on  Fixed  Beams. — Let  AB 
(Fig.  92)  represent  a  fixed  beam.     The  supports  being  fixed,  any  load  on 
the  beam,  as  is  seen,  will  produce  tension  in  the 
top  elements  of  the  beam  and  compression  in  the 
bottom  elements  of  the  beam  at  the  supports, 
while  the  reverse  will  be  the  case  out  some  dis- 
tance from  the  supports.     Then,  evidently,  there 
Fig  92  will    be   two    points    of    contra-flexure    and    the 

bending  of  the  beam  due  to  loads  supported  will 
be  as  indicated  in  Fig.  92,  where  S'  and  S  are  the  points  of  contra-flexure. 
It  is  evident  that  the  parts  AS'  and  SB  are  cantilevers,  while  the  part 
S'S  is  a  simple  beam.  In  the  case  of  fixed  beams,  the  point  of  applica- 
tion of  the  reactions  cannot  be  definitely  fixed,  so  we  deal  with  the  end 
shears  instead  of  the  reactions. 

As  a  case  of  concentrated  loads  let  P  represent  a  single  load  upon 
the  beam  at  any  distance  kL  from  A.  Now,  when  we  proceed  to  deter- 
mine the  shears  and  bending  moments  on  the  beam  due  to  this  load  P,  we 
quickly  realize  that  we  have  but  little  to  start  with,  and  it  is  only  through 
the  application  of  the  general  differential  equation  of  the  elastic  curve 
(K)  that  we  are  able  to  obtain  either.  Let  V'  and  V  represent  the  end 
shears  at  A  and  B,  respectively,  due  to  the  load  P,  and  let  M'  and  M" 
represent  the  bending  moments  at  A  and  B,  respectively.  Take  A  as  the 
origin  and  let  x  and  y  represent  the  co-ordinates  to  the  elastic  curve. 
Then,  according  to  Art.  66,  the  bending  moment  at  any  point  to  the  left 
of  the  load  any  distance  x  from  A  is 

M  =  A 
Therefore,  we  have 

~dx*~ 
and  integrating  once,  we  have 


J-i  M.  ,  *'.".         W         |  |        V        • 

dx  2 

But  dy/dx  =  0  when  x  =  0 ;  therefore,  C"  =  0,  and  we  have 


Integrating  this  equation,  we  have 

M'x-      V'x* 


THEOEETICAL  TREATMENT  OF  BEAMS  91 

But  here  y  —  0  when  x  —  0  ;  therefore,  C"  =  0,  and  we  have 
M'x2      V'x* 


Now  the  bending  moment  at  any  point  to  the  right  of  the  load 
distance  from  A  is 

Therefore, 

and  integrating  this  equation,  we  have 


E  I        =  M'  +  V'x  -Px  +  PkL, 
ax 


Now,  dy/dx  -  0  when  x  =  L.     So  substituting  L  for  x  and  equating,  we 
have 


Then  substituting  this  value  of  Cx  in  the  last  equation,  we  have 


Integrating  (3),  we  have 


V'x*     Px*     PkLx2  V'L2x     PL2* 

_    ._     __  _  -M'Lx 


But  y  -  0  when  x  =  L.     So,  substituting  L  for  x  in  the  last  equation, 
equating  and  reducing,  we  have 


M'L2      V'L*     PL*      PkL* 

~~     ~~      ~~     ~~ 


Then  substituting  this  value  of  C2  in  the  last  equation,  we  have 

M'x2      V'x*     Px*      PkLx2      __,r       V'L*x 

___  +  _     __4____M'L,--^+ 

M'L2     V'L*     PL*     PkL* 


..............  (4). 

Now  it  is  evident  that  equations  (1)  and  (3)  are  equal  when  x  =  kL 
in  each.  Then  by  substituting  kL  for  x  in  each,  equating  and  reducing, 
we  have 


Then  by  substituting  this  value  of  M'  in  each  of  the  equations  (2) 
and  (4)  and  then  substituting  kL  for  x  in  each,  we  have  the  two  equal. 
Then  by  equating  these  equations  each  to  each  and  reducing,  we  have 

V  =  P(l  +  2&8  -  3fc2)  ................................  (6). 


92  STRUCTURAL  ENGINEERING 

By  substituting  this  value  of  V  in  (5),  and  reducing,  we  have 

M'  =  PL(2k2  -  P  -  /c)  ................................  (7). 

Then,  having  V  and  M'  determined  from  (6)  and  (7)  for  the  point 
A,  the  bending  moment  at  any  other  point  in  the  beam,  due  to  the  load  P, 
can  be  determined  according  to  Art.  66,  and  the  shear  at  any  point  can 
be  determined,  as  it  is  equal  to  the  summation  of  the  forces  summed  up 
from  A  beginning  with  V  '. 

If  there  be  more  than  one  load  on  the  beam,  the  shear  and  bending 
moment  at  A  due  to  each  can  be  determined  separately  from  (6)  and  (7). 
Then  adding  these  shears  together  and  these  bending  moments  together, 
we  obtain  the  shear  and  bending  moment  at  A  due  to  all  of  the  loads. 
Then  the  bending  moment  and  shear  at  any  point  of  the  beam,  due  to  all 
of  the  loads,  can  be  determined.  That  is,  if  the  shear  and  bending 
moment  at  one  support  of  a  fixed  beam  are  known,  the  shear  and  bending 
moment  at  any  other  point  can  be  determined,  and  hence  the  equations 
(6)  and  (7)  are  all  that  are  absolutely  necessary.  But  from  these 
equations  we  can  readily  derive  equations  for  the  shear  and  bending 
moment  at  the  other  support,  if  such  be  desired.  For  example,  the  shear 
at  B,  Fig.  92,  is  equal  to  the  load  P,  producing  the  shear,  minus  the 
shear  at  A.  Then  subtracting  the  value  of  V  ,  given  in  (6),  from  P,  we 
have 


and  reducing,  we  have 

V  =  P(2P  -  3k2)  .........................  .  .  .  .  .......  (8). 

The  bending  moment  at  B  (Fig.  92)  is 

M"  =  AT  +  V'L  -  P(L  -  jfcL). 

Then  by  substituting  the  value  of   V   and  M'   given  in   (6)   and   (7), 
respectively,  we  have 

M"  =  PL(2k2-k*-k)+PL(l+2kz-3k2)-P(L-kL), 
and  reducing,  we  have 

M"  =  PL(k3  -  F)  .  .  .....................  ............  (9). 

As  an  example  of  application,  suppose  the  load  P  (Fig.  92)  to  be  at 
mid-span.  Then  k  =  1/2.  Now  substituting  this  1/2  for  k  in  (6)  and 
(7),  we  have 


- 


respectively,  which  is  the   shear  and   bending  moment  at  A.      For  the 
bending  moment  under  the  load  we  have 


THEOEETICAL  TREATMENT  OF  BEAMS  93 

In  case  a  uniform  load  extends  over  the  entire  length  of  a  fixed  beam, 
the  end  shears  are  each  equal  to  one-half  of  the  total  load  on  the  beam. 
Suppose  the  beam'  shown  in  Fig.  92  supports  a  uniform  load  of  w  pounds 
per  foot  of  length  extending  over  the  entire  span.  Let  V  and  Vff  repre- 
sent the  end  shears  at  A  and  B,  respectively,  due  to  this  uniform  load, 
and  let  M'  and  M"  represent  the  bending  moments  at  A  and  Bf  respect- 
ively, due  to  the  same. 

Taking  A  as  the  origin  as  before,  we  have  for  the  bending  moment 
at  any  point  x  distance  from  A 


therefore, 


Integrating  this  equation,  we  have 


i. 

dx  46 

But  dy/dx  =  0  when  x  =  0  ;  therefore,  Cl  -  0,  and  we  have 


Integrating  again,  we  have 

M'x2      wLx3      wx* 


But  y  =  0  when  x  -  0  ;  therefore,  C2  =  0,  and  we  have 


a 


But  y  —  0  also  when  x  =  L.     Then  substituting  L  for  x  in  (11),  and 
reducing,  we  have 


which  is  the  bending  moment  at  A.  Then  knowing  the  bending  moment 
and  shear  at  A,  the  bending  moment  and  shear  at  any  other  point  in  the 
beam  can  be  determined.  For  example,  the  bending  moment  at  the 
center  is 


To  determine  the  points  of  contra-flexure,  we  simply  write  the 
equation  for  the  bending  moment  at  any  point  and  equate  it  to  zero.  For 
example,  the  bending  moment  at  any  point  x  distance  from  A,  when  the 
beam  supports  a  uniform  load  of  w  pounds  per  foot  (Fig.  92),  is 

wx2         wL2     wLx     wx* 

M=^+rx___=__+__  __. 


94 


STRUCTURAL  ENGINEERING 


Then  by  equating  this  to  zero,  we  have 


w  x- 

~ 


and  solving  for  x,  we  have 
*  =  -  ± 

/v 


=  0.21  Lor  0.79  L  (about). 


That  is,  one  point  of  contra-flexure  is  0.21  of  L  from  A  —  considered 
in  practice  as  being  J  of  L  —  and  the  other  0.79  of  L  from  A  —  considered 
in  practice  as  being  J  of  L.  The  points  of  contra-flexure  can  be  deter- 
mined in  a  similar  manner  in  case  of  any  other  kind  of  loading. 

Deflections,  if  desired,  can  be  computed  from  equations  (2),  (4), 
and  (11). 

70.    Bending  Moments,  Shears,  and  Reactions  on  Continuous 

Beams.  —  Let  ABC  (Fig.  93)  represent  two  consecutive  spans  of  a  con- 
tinuous beam  of  n  spans.     The  bending  or  curving  of  the  beam  in  the  two 

J=L. 


Fig.  93 

spans  is  shown  to  be  similar  to  that  of  two  fixed  beams,  each  span  being 
considered  a  beam.  If  the  spans  were  equal  in  length  and  symmetrically 
loaded,  this  would  practically  be  the  case,  but  otherwise  it  would  not  be, 
for  it  is  readily  seen  that  the  loads  in  one  span  would  tend  to  produce 
reverse  bending  in  the  adjacent  span,  which  would  curve  it  up  instead 
of  down.  For  example,  the  loads  in  span  AB  could  be  such  that  the  span 
BC  would  be  curved  upward  instead  of  downward,  as  it  is  shown.  So  it  is 
evident  that  the  tangent  to  the  elastic  curve  at  any  support  is  horizontal 
only  when  the  spans  are  equal  in  length  and  rigidity,  and  symmetrically 
loaded.  Hence  the  slope  of  the  elastic  curve  at  the  supports  cannot,  in 
general,  be  utilized  in  preliminary  investigations,  as  in  the  case  of  fixed 
beams. 

In  the  following  treatment  it  is  assumed  that  the  beams  have  a 
uniform  cross-section  and  are  homogeneous  throughout,  and  are  supported 
upon  simple  supports  of  equal  elevation.  This  is  an  assumption  usually 
made  in  practice.  The  equation  employed  in  analyzing  continuous  beams 
is  known  as  the  "Three-Moment  Equation/'  which  we  shall  now  proceed 
to  derive. 

Case  I.     When  the  beam  supports  concentrated  loads. 

Referring  to  Fig.  93  let  L  be  the  length  of  the  span  AB  and  L,  the 
length  of  the  span  BC.  Let  P  represent  a  load  at  any  point  kL  distance 
from  A  in  span  AB  and  let  P'  represent  a  load  at  any  point  k,L,  distance 
from  B  in  span  BC.  Let  M' ',  M" ,  and  AT"  represent  the  bending 


THEOEETICAL  TEEATMENT  OF  BEAMS  95 

moments  at  the  supports  A,  B,  and  C,  respectively,  due  to  the  loads  P 
and  P',  and  let  V  and  V  represent  the  shears  just  to  the  right  of  the 
supports  A  and  B,  respectively,  due  to  these  same  loads. 

Considering  the  span  AB,  and  taking  A  as  the  origin,  the  bending 
moment  (according  to  Art.  66)  at  any  point  to  the  left  of  the  load  P,  x 
distance  from  A,  is 

therefore, 

Integrating  once,  we  have 

EI  £ = M'x  +ir +  c' (1)- 

Integrating  again,  we  have 


But  y  —  0  when  x  =  0  ;  therefore,  C"  —  0,  and  we  have 

M'r2        T7r3 

*-+C'i  ..............................  (2). 


Now,  the  bending  moment  at  any  point  to  the  right  of  the  load  P, 
x  distance  from  A,  is 

therefore, 

Integrating  once,  we  have 


El        =  AP  +  Vx  -  Px  +  PkL. 

dx2 


(3). 


Integrating  again,  we  have 

M'x2      Vxz     Pxz     PkLx2 
EIy  =  ~^--  +  —  --g-  +  —  g—  +  C/ar  +  C  ...............  (4). 

It  is  readily  seen  that  equations  (1)  and  (3)  are  equal  when  x-'kL 
in  each.     Then  substituting  kL  for  x  in  each  and  equating,  we  have 


6  6 

from  which  we  obtain 


Now  substituting  this  value  of  C'  in  (2),  we  have 
M'x*      Vx*     P 


96  STRUCTURAL  ENGINEERING 

Now  in  equation   (4)  y  =  0  when  x  =  L.     Then  substituting  L  for  x 
in  that  equation,  we  have 

M'L2      FL3__PL3_     PkL3 


from  which  we  obtain 


M'i*     rL*    PL*     PkL* 

~~    ~~ 


' 

Substituting  this  value  of  C,,  in  equation  (4),  we  have 
M'x2      Fx3      Px3     PkLx2 


M'L2     FL3     PL3     PkL3 


IT 


Equation  (6)  applies  to  the  part  of  the  elastic  curve  to  the  left  of  the 
load  P,  while  (8)  applies  to  the  part  of  the  curve  to  the  right.  Then  these 
two  equations  are  equal  when  x  =  kL  in  each.  So  substituting  kL  for  x  in 
each,  equating  and  reducing,  we  have 

r  _    Pk3L2     M'L     FL2     PL2     PkL2 

6  2          6       ~6~     ~T~ 

Then  substituting  this  value  of  C,  in  (5),  we  have 

„,     Pk*L*     Pk3L2     M'L     FL2     PL2     PkL2 

"=—2~   -IT  TT  Tr*nr  "T" 

and  substituting  the  value  of  C,  given  by  (9)  in  (7)  and  reducing,  we  have 


So  we  have  thus  determined  all  of  the  constants  of  integration  so  far 
involved. 

Now  substituting  in  (1)  the  value  of  C"  given  in  (10),  we  have 

Fj^lL^M'       —  PPL2     PM2     M/L 

dx~           '2  2              6             2 

FL2     PL*  PkL2 

+ 


662 
and  substituting  the  same  value  of  C'  in  (2),  we  have 

M'x2      vx3  ^  Pk2L2x  _  Pk3L2x  _  MfLx 
2      +   6  2  ~T~  2 


PkL2x 

6     +—6 2~ W' 

Next  substituting  in  (3)  and  (4)  the  value  of  C,  and  C,, ,  respectively, 
given  in  (9)  and  (11),  we  have 


THEOKETICAL  TKEATMENT  OF  BEAMS  97 

,Tdy                   Vx2      Px2                     PksL2 
EI-^=M'x  +  - —  +  PkLx — 

dx  22  6 

-¥-?**?-*?• («> 

and 

_M'x2      Vx3     Px3      PkLx2      Pk*L2x 
EIy~~2~    '~Q~     ~T       ~T~        ~6~ 

M'Lx     VL2x      PL2x      PkL2x 

+ 


26626 


Equations  (12)  and  (13)  apply  to  the  part  of  the  elastic  curve  to  the 
left  of  the  load  P,  while  equations  (14)  and  (15)  apply  to  the  part  of  the 
elastic  curve  to  the  right  of  the  load  P.  Now  it  is  evident  that  by  pro- 
ceeding in  the  same  manner  with  span  BC,  four  equations  corresponding 
to  (12),  (13),  (14),  and  (15)  could  be  derived  for  the  elastic  curve  in 
that  span.  But  it  is  readily  seen  that  the  equations  would  differ  from  the 
above  equations  only  in  the  marking  of  the  M's,  V's,  P's,  k's,  and  Us.  So 
for  span  BC,  we  can  write 


dx  226 

_  M"L,  _  V'L,2     PfL2     P'k,L,2 

2662 


(16). 


6  O  4>  O 

M"L,x     V'L,2x      P'L,2x      P'lc,L*x 


2662 
for  the  part  of  the  elastic  curve  to  the  left  of  the  load  P'  ,  and 

F7dy-M»XJ'X*     P'X*     P'kLx     P'*'L> 
(  J&4*  ~  ~2~      ~^~  '    '          ~Mp 

M"L,     V'L;<     P'L2     P'k,L,2 


(17) 


M"x2     V'x*     P'x*     P'k,L,x2     P'Jc,3L,"x      M"L,x 

~Y~     ~6~     ~6~        ~T~  ~Q~          ~~2~ 

V'L,*x     P'L,2x      P'k,L,2x     P'k*L,s 
+  —  «  ---  H  -  +  -  - 


«  « 

6  o  2 

for  the  part  of  the  elastic  curve  to  the  right  of  the  load  P'. 

Now  taking  moments  about  B,  according  to  Art.  66,  we  have 


from  which  we  obtain 


98  STRUCTURAL  ENGINEERING 

Then  substituting  this  value  of  V  in  (14)  and  reducing,  we  have 
^dy  M"x2     M'x2     Pkx2  Pk*L* 

EIfx=Mx+^r-*r-  — 

M'L     M"L     M'L     PkL2 

2  6         ~~6~      ~T~ 

Likewise,  taking  moments  about  C,  we  have 

M'"  =  M"+F'L,-P'  (!,, 
from  which  we  obtain 

M  '  "  -  M" 


(22). 


Then  substituting  this  value  of  V  in  (16),  we  have 


(23). 
2662 

Equation  (21)  applies  to  the  part  of  the  elastic  curve  to  the  right  of 
the  load  P  in  span  AB,  and  equation  (23)  applies  to  the  part  of  the 
elastic  curve  to  the  left  of  the  load  P'  in  span  BC.  Now  it  is  readily  seen 
that  these  two  equations  are  equal  when  x  —  L  in  (21)  and  0  in  (23),  as 
the  two  curves  have  a  common  tangent  at  B.  Then  substituting  L  for  x 
in  (21),  and  0  for  x  in  (23),  equating  and  reducing,  we  have 


M"L     M'L,     PkL2     Pk*L2        M"L,     M 


3           6            6  6  3              6 

P'k,2L,2  P'k,3L,2  P'k,L,2 

"T~  "6"  I" 
from  which  we  obtain 

M'L  +  2M"(L  +  L,)+M"'L,=  -PL2(k-kz)-P'L,2(2k,-3k,2  +  k*)  .  (L), 

which  is  the  three-moment  equation  when  the  loads  are  concentrated  loads. 
The  more  general  equation,  as  usually  written,  is 


Case  II.     When  the  beam  supports  uniform  loads. 

Let  w  be  the  uniform  load  per  foot  on  span  AB  (Fig.  93)  and  let  wf 
be  the  uniform  load  per  foot  in  span  BC.  Let  M'  ,  M"  ,  and  M'  "  now 
represent  the  bending  moments  at  the  supports  A,  B,  and  C,  respectively, 
due  to  the  above  uniform  loads,  and  let  V  and  V  now  represent  th*- 
shears  just  to  the  right  of  supports  A  and  B,  respectively,  due  to  these 
same  loads.  Considering  span  AB,  and  taking  A  as  the  origin,  the  bend- 
ing moment  at  any  point  x  distance  from  A  is 


THEORETICAL  TREATMENT  OF  BEAMS  99 

therefore, 


Integrating  once,  we  have 

Eii=M'^F- 

Integrating  again,  we  have 


M'x2     Fx5      wx 


6         M 
But  y  —  0  when  x  —  0  ;  therefore,  C"  =  0,  and  we  have 

M'x2      Fx*     wx* 


But  also  y  =  0  when  x  —  L.     Then  substituting  L  for  x  in  the  last 
equation,  we  have 

ML*     FL* 

-r-+^r- 

from  which  we  obtain 

M'L     F 


Substituting  this  value  of  C  in  equation  (24),  we  have 


Similarly,  if  the  origin  be  taken  at  the  support  B,  for  span  BC,  we  have 
for  the  bending  moment  at  any  point  x  distance  from  B 


therefore, 

EI dx~< 


Integrating  once,  we  have 

£7^=M",  +  ^-^  +  C  ........................  (36). 

dx  26 

Integrating  again,  we  have 

M"x2      V'x*     w'x* 

-^r+-6---w+ 

But  x  -  0  when  y  -  0  ;  therefore,  C,,  =  0,  and  we  have 
Af"a?2     F'as     w'a;4     „ 

"^      " 


But  y  =  0  when  a?  =  L,  .     Then  substituting  L,  for  j7  in  the  last 
equation,  we  have 


100  STRUCTURAL  ENGINEERING 

wi*    VL*    ™L* 


2  6  2 

from  which  we  obtain 

M"L,     V'L?     w'L? 
2          ~6~     ~W 
Now  substituting  this  value  of  C,  in  equation  (26),  we  have 


..    , 

dx  2  6  2  fi  24 

It  is  readily  seen  that  equations  (25)  and  (27)  would  be  equal 
when  x  =  L  in  (25)  and  x  -  0  in  (27),  the  two  elastic  curves  having  a 
common  tangent  at  B.  Then  substituting  L  for  x  in  (25)  and  0  for  x  in 
(27),  equating  and  reducing,  we  have 

12M'L  +  8FL2  -  3wL5  =  -12M"L,  -  4F'L,2  +  w'L,s  .....  (28). 

Now  (referring  to  span  AB)  taking  moments  about  the  support  B, 
we  have,  according  to  Art.  66, 


Transposing  and  dividing  by  L,  we  have 
M»-M>     wL 

L  2 C     '* 

Next  (referring  to  span  BC)  taking  moments  about  C,  we  have 

M'  "  =  M"  +  V'L,  -  - — '-  • 
Transposing  and  dividing  by  L, ,  we  have 


Now  substituting  the  values  of  V  and  V  as  given  by  (29)  and  (30), 
respectively,  in  equation  (28),  and  reducing,  we  have 

M'L  +  2M"(L  +  L,)  +  M'"L,  =  _!!_  ^! 


which    is    the    three-moment    equation    when    the    loads    are    uniformly 
distributed. 

The  three-moment  equation,  as  is  readily  seen,  is  an  expression  for 
the  bending  moments  at  any  three  consecutive  supports  in  terms  of  the 


M' 


Fig.   94  Fig.   95 

lengths  of  the  two  intervening  spans  and  the  loads  supported  in  these 
spans.  The  lengths  and  loads,  of  course,  are  known  quantities.  Now 
for  any  continuous  beam  of  n  spans,  there  are  always  (n  +  1)  supports, 


THEOEETICAL  TEEATMENT  OF  BEAMS  101 

and  for  each  three  consecutive  supports  a  three-moment  equation  can  be 
written.  Then,  (n  —  1)  three-moment  equations  can  be  written  for  any 
continuous  beam.  For  example,  let  abcdef  (Fig.  94)  represent  a  con- 
tinuous beam  of  five  spans.  Here  we  can  write  four  (which  is  (n  —  1)) 
three-moment  equations:  one  for  supports  a,  b,  and  c;  one  for  supports 
b,  c,  and  d;  one  for  supports  c,  d,  and  e;  and  one  for  supports  d,  e,  and  /. 
Now  the  bending  moment  at  each  of  the  supports  would  be  included  in 
these  four  three-moment  equations,  being  (n  -f  1)  moments  in  all.  But, 
as  the  bending  moments  at  both  a  and  /  are  equal  to  zero,  the  four 
unknown  moments  at  b,  c,  d,  and  e  can  be  determined,  as  we  have  four 
equations  and  four  unknown  moments.  Then,  after  these  unknown 
moments  at  the  supports  are  determined,  the  reactions  at  each  support 
can  be  determined  and  then  the  shear  and  the  bending  moment  at  any 
point  of  the  beam  can  be  determined. 

As  an  example,  let  ABC  (Fig.  95)  represent  a  continuous  beam 
having  three  supports  which  would  be  known  as  a  beam  continuous  over 
three  supports.  Let  M'  ',  M"  ,  and  M'  "  represent  the  bending  moments 
at  the  supports  A,  B,  and  C,  respectively,  due  to  any  loads  that  we  choose 
to  consider,  and  let  R\9  R2,  and  R3  represent  the  reactions  at  the  sup- 
ports A,  B,  and  C,  respectively,  due  also  to  any  loads  that  we  choose  to 
consider.  The  intensity  of  these  moments  and  reactions  would,  of  course, 
vary  with  the  loads. 

Let  us  first  consider  two  concentrated  loads  P  and  P'  on  the  beam  as 
shown  in  Fig.  95.  Now,  as  there  are  but  three  supports,  only  one  three- 
moment  equation  can  be  written.  So  writing  this  one,  which  is  really 
Formula  (L),  given  on  page  98,  we  have 

M'L  +  %M"(L  +  L,)  +  M'  "L,  =  -PL2(k  -  P)  -  P'L,2(2k,  -  3k,2  +  k,3). 
But  in  this  case  M'  =  0  and  M'  "  —  0.     So  the  equation  becomes 
2M"(L  +  L,)  =  -PL2(k  -  P)  -  P'L2(2k,  -  3k2  +  A;,3), 

from  which  we  obtain 

PL*(k-k*)-P'L?(2k,-3k;-  +  k,°) 


Now   considering  points   A    and  B,   and  taking   moments   about  B, 
according  to  Art.  66,  we  have 

M"=RlL-P(L-kL). 
Transposing  and  reducing,  we  have 


Now,  by  substituting  the  value  of  M"  ',  as  given  in  equation  (31),  in 
this  last  equation,  the  value  of  the  reaction  R1  at  A  is  obtained.  Then, 
after  Rl  is  known,  the  bending  moment  at  any  point  in  the  span  AB  is 
obtained  as  readily  as  in  the  case  of  a  simple  beam,  as  it  is  equal  to  the 
algebraic  sum  of  the  moments  about  the  point  considered  of  the  forces  to 
the  left  of  the  point,  which  are  now  all  known. 


102  STRUCTURAL  ENGINEERING 

Now,  similarly,  considering  points  B  and  C,  and  taking  moments 
about  B,  we  have 


Transposing  and  reducing,  we  have 


Then  by  substituting  the  value  of  M"  ,  as  given  in  equation  (31),  in  this 
last  equation,  the  value  of  the  reaction  R3  at  C  is  obtained.  Then  after 
R3  is  known,  the  bending  moment  at  any  point  in  the  span  BC  is  readily 
obtained,  as  it  is  equal  to  the  algebraic  sum  of  the  moments  about  the 
point  considered  of  the  forces  to  the  right  of  it,  which  are  now  all  known. 
The  shear  at  any  point  in  either  span  is  obtained  by  simply  adding  up 
(algebraically  in  each  case)  the  forces  between  the  end  and  the  point 
considered.  R2,  as  is  readily  seen,  is  equal  to  the  sum  of  the  loads  minus 
the  two  reactions  R\  and  R3.  It  is  also  equal  to  the  shear  just  to  the 
right  of  B  plus  the  shear  just  to  the  left  of  B.  By  taking  moments  about 
both  A  and  B,  equations  can  be  derived  from  which  these  shears  can  be 
obtained,  and  consequently  R2. 

Now  it  is  evident  that  the  bending  moments,  shears,  and  reactions  on 
the  above  beam  due  to  any  number  of  such  loads  could  be  determined  by 
considering  the  loads  in  pairs,  as  in  the  above  case.  But  it  is  more  con- 
venient to  derive  equations  expressing  the  values  of  these  for  one  single 
load  at  any  point.  So  let  P  (Fig.  95)  be  the  only  load  on  the  beam. 
Then  Pf  and  k,  will  not  occur  in  the  three-moment  equation,  that  is,  they 
will  be  equal  to  zero,  and  hence  the  three-moment  equation  becomes 


from  which  we  obtain 
* 
' 


Now  taking  moments  about  B,  considering  points  A  and  B,  we  have 
M"  =  LRl-P(L-kL). 

Then  substituting  this  value  of  M"  ,  as  given  in  (33),  in  this  last  equation, 
and  reducing,  we  have 

..........................  (34)- 


Then  taking  moments  about  B,  considering  points  B  and  C,  we  have 


Substituting  this  value  of  M",  as  given  in  (33),  in  this  last  equation,  and 
dividing  by  L,  ,  we  have 

'          PL'(k-V) 

-ZL,(L  +  L.)  .................................  (3' 

Now,  Rl  and  R3  can  be  determined  for  any  load  in  the  span  AB 
from  equations  (34)  and  (35),  respectively,  and  likewise  for  any  load  in 


THEOEETICAL  TEEATMENT  OF  BEAMS  103 

span  EC  by  interchanging  the  R's,  that  is,  #1  would  in  that  case  be  at  C 
and  R3  at  A  —  or  just  turn  the  beam  end  for  end.  If  there  are  several 
loads  in  the  two  spans,  the  reactions  R\  and  R3  due  to  each  can  be  deter- 
mined separately  and  added  algebraically  (as  R3  is  always  minus),  and 
thus  the  total  reaction  at  both  A  and  C  would  be  determined.  Then  the 
bending  moment  and  shear  at  any  point  in  the  beam  is  readily  determined, 
as  explained  above. 

Now  if  the  two  spans  are  equal,  we  have  L  =  L,  in  both  equations 
(34)  and  (35).  Then  (34)  reduces  to 

Rl  =  :  ?  (4  -  5fc  +  P)  ................................  (36), 

and  (35)  reduces  to 

R3  =  -  ?  (k  -  P)  ..................................  (37). 

Then  adding  (36)  and  (37)  and  subtracting  the  result  from  the  load  P 
and  reducing,  we  have  the  formula 

R2  =?•  (3k  -  P)  ....................................  (38). 

</ 

Now  the  reactions  due  to  a  load  at  any  point  on  any  beam  continuous 
over  three  supports  and  having  equal  spans,  can  be  determined  from 
equations  (36),  (37),  and  (38).  Then  the  bending  moments  and  shears 
are  readily  determined  as  explained  above. 

As  a  practical  example,  let  L  and  L,  in  Fig.  95,  each  be  equal  to  12 
feet,  and  let  P  be  4  feet  from  A  and  let  P'  be  6  feet  from  C.  Then 
k  =  4/12  =  1/3,  and  k,  =  6/12  =  1/2. 

First  considering  P  alone  and  substituting  1/3  for  k  in  (36)  and 
reducing,  we  have 


for  the  value  of  the  reaction  at  A  due  to  the  load  P.     Next  substituting 
1/3  for  k  in  (37)  and  reducing,  we  have 


for  the  reaction  at  C,  which  is  minus  ;  that  is,  it  pulls  down  upon  the  beam 
instead  of  pushing  up.  Now  after  these  reactions  are  determined,  the 
bending  moment  and  shear  at  any  point  in  the  beam  due  to  the  load  P 
alone  can  be  determined  as  explained  above.  The  bending  moments  in 
span  EC  would  be  negative,  as  #3  is  negative;  that  is,  the  top  elements  in 
the  beam  would  be  in  tension  while  the  bottom  ones  would  be  in  com- 
pression. 

Now  considering  P'  alone,  and  substituting  1/2  for  k  (  =  &,)  in  (36) 
and  reducing,  we  have 


104  STRUCTURAL  ENGINEERING 

which  is  the  reaction  at  C  due  to  load  P'  in  span  BC.    Next,  substituting 
1/2  for  k  in  (37)  and  reducing,  we  have 


which  is  the  reaction  at  A  due  to  the  load  P'  in  span  BC.     Then  the 
reaction  at  A  due  to  the  two  loads  is 


_ 

27         32     ' 

and  the  reaction  at  C  due  to  the  same  is 

13         13 
32         54 

After  these  reactions  are  thus  determined,  the  bending  moment  and  shear 
at  any  point  in  the  beam  are  readily  determined,  as  explained  above. 

Now  instead  of  the  concentrated  loads  just  considered,  suppose  the 
beam  shown  in  Fig.  95  supported  a  uniform  load  of  w  pounds  per  foot  in 
span  AB  and  w'  pounds  per  foot  in  span  BC,  extending  over  the  entire 
span  in  each  case.  In  this  case,  as  the  loads  are  uniformly  distributed, 
we  would  use  the  three-moment  equation  (M),  given  above,  which,  as  Mf 
and  M"  are  equal  to  zero,  reduces  to 

8M"<L  +  L,)=--  .........................  (39). 


Then  transposing  and  dividing  through  by  2(L  +  L,),  we  have 


Now  taking  moments  about  B,  and  considering  points  A  and  B,  according 
to  Art.  66,  we  have 


Then  substituting  the  value  of  M",  as  given  in  (40),  in  this  last  equation, 
transposing,  and  reducing,  we  have 

wL      wL3+  w'L*  n 

~-~-  '  ......................... 


Now  taking  moments  about  B,  and  considering  points  B  and  C,  we  have 


and  substituting  the  value  of  M",  as  given  in  (40),  in  this  last  equation, 
transposing  and  reducing,  we  have 


After  these  reactions  are  obtained,  the  bending  moment  and  shear  at  any 

point  in  the  beam  can  be  determined  very  readily  as  explained  above. 

Suppose  the  spans  to  be  equal  in  length  and  suppose  the  loads  are 


THEORETICAL  TREATMENT  OF  BEAMS  105 

equal.  Then  L  =  L,  and  w  =  w'  ,  and  substituting  L  for  L,  and  w  for  w' 
in  both  (41)  and  (42),  and  reducing,  we  have 

3  =  ?-wL, 

o 

that  is,  the  reactions  at  A  and  C  are  equal  to  f  of  the  load  on  one  span. 
Then,  evidently,  the  shear  just  to  the  right  and  also  just  to  the  left  of  B 
is  equal  to  f  (wL}  or  f  of  the  load  on  one  span,  and  hence  for  the  reaction 
at  B  we  have 


Substituting  L  for  L,  and  w  for  w'  in  (39)  and  reducing,  we  have 

' 


which  is  the  bending  moment  at  B,  the  central  support.  This  being  minus 
shows  that  the  top  elements  of  the  beam  are  in  tension  while  the  bottom 
ones  are  in  compression.  This  same  moment  can  be  obtained  by  taking 
moments  about  B  and  considering  either  span,  say,  span  AB.  Then  we 
have 


But  Rl  =  f  (te>L),  as  shown  above.     So  substituting  this  value  for  Rl  in 
this  last  equation,  and  reducing,  we  have 


Now  taking  moments  about  the  center  of  span  AB,  we  have 

Substituting  f  (wL)  for  Rl,  we  have  for  the  bending  moment  at  mid-span 


which  is  one-half  of  what  it  is  at  B,  the  central  support. 

Beams  having  four  or  more  supports  can  be  analyzed  in  a  similar 
manner.  Of  course,  there  would  be  more  three-moment  equations  involved 
— being  (w  —  1)  in  each  case.  However,  the  same  general  method  as 
outlined  above  would  hold  for  all  cases. 


CHAPTER  V 


THEORETICAL  TREATMENT  OF  COLUMNS 


L_ 


71.  Preliminary.— Let  CB  (Fig.  96)  represent  a  round  steel  rod 
of  uniform  cross-section  A  and  length  L  standing  vertically  upon  a 
smooth,  rigid  base  at  B  and  supporting  a  load  P  at  C  which  is  sym- 
metrically applied  in  reference  to  the  longitudinal  axis  of 
the  rod ;  that  is,  the  load  is  applied  in  the  center  of  gravity 
of  the  cross-section  of  the  rod.  It  is  evident  that  the  load 
P  will  produce  a  direct  simple  compressive  stress  P  at 
every  cross-section  of  the  rod  throughout  its  length,  and 
hence  the  direct  compressive  unit  stress  at  every  section 
will  be  equal  to  P/A.  All  bodies  having  a  length  much 
greater  than  their  width,  as  the  above  rod,  and  loaded  in 
the  same  manner,  thereby  being  in  compression,  are  known 
in  general  as  columns. 

Now  suppose  the  above  rod  to  be  2  inches  in  diameter 
and,  say,  4  inches  long.  Then  the  unit  compressive  stress 
would  be  equal  to 

P  P 


R 
Fig.  96 


7T 


which  corresponds  to  P/A  given  above.  It  is  evident  that  this  short  rod 
would  support  a  very  heavy  load  without  failure,  in  fact,  the  compressive 
stress  P/TT  could  be  as  great  as  if  it  were  a  tensile  stress. 
But  suppose  the  rod  to  be  20  feet  long  instead  of  4  inches. 
We  know  that  the  rod,  if  20  feet  long,  would  not  support  as 
heavy  a  load  as  it  would  if  only  4  inches  long,  as  the  longer 
rod  would  be  too  limber,  that  is,  it  would  fail  by  bending 
transversely;  of  course,  the  direct  compressive  stress  would 
be  acting  just  the  same  in  either  case. 

If  a  column  were  absolutely  straight  and  the  load 
applied  absolutely  in  the  longitudinal  axis,  the  bending  re- 
ferred to  above  would  not  occur.  But  such  conditions  do  not 
exist,  as  we  know  from  experience.  So  in  the  designing  of 
columns  the  stress  due  to  this  bending  must  be  taken  into 
account  in  addition  to  the  direct  compressive  stress. 

72.  Rankine's  Formula.— Let  ED  (Fig.  97)  repre- 
sent a  column  which  bends  as  indicated  under  a  load  P.  It 
is  evident  that  the  elements  on  the  concave  side  of  the  column 
are  in  compression,  owing  to  the  bending,  while  the  elements 
on  the  other  side  are  in  tension,  exactly  as  in  the  case  of  a 
loaded  beam.  Now  it  is  readily  seen  that  the  maximum  stress 

106 


Fig.  07 


THEOEETICAL  TREATMENT  OF  COLUMNS         107 

on  the  column  will  occur  on  the  compression  or  concave  side,  for  here  the 
direct  and  bending  stresses  combine,  while  the  tensile  stress  on  the  other 
side  reduces  the  direct  stress.  This  maximum  compressive  stress  will,  as 
is  readily  observed,  occur  at  the  middle  of  the  column,  where  the  deflection 
is  greatest,  which  is  represented  as  A.  The  bending  moment  on  the 
column  at  any  point  x  distance  from  E  due  to  the  load  is 

M  =  Pz, 

where  z  is  the  deflection.  This  moment  will  be  a  maximum  when  z  =  A. 
Then  we  have 

M  =  PA 

for  the  maximum  bending  moment  on  the  column.  But  from  (D),  Art. 
53,  we  have  /  =  My  /I,  and  substituting  PA  for  M,  we  have 


(where  y  =  the  distance  from  neutral  axis  of  column  to  the  extreme 
elements)  for  the  maximum  compressive  stress  on  the  column  due  to 
bending.  Now  if  we  add  this  to  the  direct  stress,  P/A,  we  have 

P 


which  is  the  maximum  compression  unit  stress  on  the  column. 

Now  —  as  is  proven  by  experiment  —  A  will  vary  directly  as  L2  and 
inversely  as  y;  that  is, 


y 

Then  if  L2/y  be  multiplied  by  some  constant  c,  we  have 

L2 
A=  c  -- 

y 

Substituting  this  value  of  A  in  (1),  we  have 

P     PcL2 


But  7  =  Ar2   (Art.  49).     Then  substituting  this  value  of  I  in  the  last 
equation  and  reducing,  we  have 


This  can  be  written  in  the  form 
P          x 

.......................................  (O), 

which  is  known  as  Rankine's  Formula,  where 

p  —  maximum  compressive  unit  stress  on  the  column; 

P  =  load  on  the  column; 

L  =  length  of  column  in  inches3' 


108  STRUCTURAL  ENGINEERING 

A  =  area  of  cross-section  of  the  column  in  square  inches ; 
r  —  least   radius   of   gyration   of   the   cross-section   in   inches   in 

reference  to  the  neutral  axis  of  the  column; 
c  =  a  constant  which  is  determined  by  experiment,  and  depends 
upon  the  material  composing  the  column  and  upon  end 
conditions. 

The  maximum  unit  stress  on  any  known  column  due  to  a  given  load 
can  be  determined  from  Formula  (N)  and  the  allowable  direct  or  average 
compressive  unit  stress  from  (O),  providing  the  proper  value  of  c  is 
known,  which  is  determined  from  experiment. 

The  following  practical  values  of  c  are  recommended: 

For  timber  columns  c  —  4/3,000; 
For  cast-iron  columns  c  —  4/5,000 ; 
For   (medium)   steel  columns  c  -  1/11,000. 
(American  Bridge  Co.) 

If  p  be  taken  as  the  elastic  limit  of  the  material  in  the  column,  P/A 
will  be  the  direct  compressive  unit  stress  that  the  column  would  be  sub- 
jected to  when  the  elastic  limit  of  the  column  was  reached,  but  if  p  be 
taken  as  the  ultimate  strength  of  the  material,  P/A  would  be  the  direct 
compressive  unit  stress  that  the  column  would  be  subjected  to  at  failure. 
If  p  be  given  either  of  these  values,  a  factor  of  safety  must  be  used  in 
determining  the  "working  stress"  for  any  column.  If  32,000  Ibs.  be 
taken  as  the  value  of  p  at  the  elastic  limit  of  steel,  a  factor  of  safety  of 
2  should  be  used,  and  if  64,000  Ibs.  be  taken  as  the  value  of  p  when  the 
column  fails,  a  factor  of  4  should  be  used.  For  practical  designing,  it  is 
more  convenient  to  take  p  as  the  allowable  or  working  stress  of  the 
material  in  tension.  This  for  medium  steel  may  be  taken  as  16,000  Ibs., 
in  accordance  with  the  specifications  of  A.  R.  E.  Ass'n.  Now,  substituting 
this  value  of  p  and  the  above  value  of  c  in  Rankine's  Formula  (O),  we 
have 

P  16,000 


11,000 

This  Formula  (P)  may  be  used  for  designing  any  steel  column,  but 
a  Straight  Line  Formula  is  preferable,  as  it  is  more  readily  applied  and 
the  results  obtained  are  practically  as  accurate. 

73.  Straight  Line  Formula.— The  ordinates  to  the  curve  abed 
(Fig.  98)  measured  from  the  horizontal  line  OH  give  the  value  of  P/A 
corresponding  to  the  different  values  of  L/r,  as  shown.  This  curve  is 
obtained  by  substituting  the  different  values  of  (L/r)2  in  Formula  (P). 
For  example,  the  ordinate  eb  is  equal  to 

P  16,000 

A~  1  +  OT^30)2"    ' 

It  is  readily  seen  from  Formula  (P)  that  the  ordinate  Oa  is  equal  to 
16,000  Ibs.,  as  L/r  =  0.  Now,  if  some  line  as  ak  be  drawn  such  that  the 
ordinates  to  it  will  be  practically  the  same  as  to  the  curve,  it  is  obvious 


THEOEETICAL  TREATMENT  OF  COLUMNS 


109 


«_     -^_    „„    —       ._     80     9O    1OO   //O  t2O 

Vafvesof  A  (zo">3Cafe) 

Fig.   98 

that  the  equation  to  this  line  can  be  used  as  a  column  formula  instead  of 
Formula  (P),  and  the  equation  to  the  line  would  be  known  as  a  "Straight 
Line  Formula." 

The  ordinate  Hk  to  this  line  is  taken  here  as  7,600  Ibs.,  and  the 
ordinate  Oa  is  16,000  Ibs.  Then,  as  OH  is  known,  the  slope  of  the  line  ak 
can  be  determined,  and  consequently  its  equation  in  reference  to  0  can  be 
derived. 

Let  6  be  the  angle  that  the  line  makes  with  the  horizontal.  Then  we 
have  any  ordinate  y,  distance  x  from  0,  given  by  the  formula 

y  =  16,000  -  .rtantf  ...................................  (1). 

Nowtan6=(Oa-Hk)  +  OH  =  0.14:,  as  0  a  and  Hk  are  equal  to  16,000  and 
7,600  Ibs.,  respectively,  and  OH,  to  the  same  scale,  is  equal  to  60,000  Ibs. 
Now  for  x,  in  terms  of  L/r,  we  have 


(Q), 


Then  substituting  these  values  of  tan0  and  x  in  (1),  we  have 


=     =  16,000  - 


which  is  the  straight  line  formula  as  given  in  the  A.  R.  E.  Ass'n  Speci- 
fications. 

As  is  obvious,  the  author  just  deliberately  took  the  ordinate  Hk  so 
that  the  above  formula  would  result.  If  the  line  ak  be  drawn  to  conform 
to  mere  observation,  a  slightly  different  formula  would,  very  likely,  be 
obtained. 

It  is  readily  seen  that  a  straight  line  formula  for  any  other  material 
can  be  determined  in  the  same  manner  by  simply  using  the  proper  c  and 
working  stress  in  each  case. 

74.    Examples  in  the  Application  of  Column  Formulas. — 

Example  1.  What  will  be  the  maximum  compressive  unit  stress 
produced  in  a  wooden  column  10  x  10  ins.  in  cross-section  and  12  ft. 
long  by  a  load  of  20,000  Ibs.? 

Here  A  =  100,  L  =  144,  r  =  2.88,  and  c  =  4/3,000.  The  value  of  p  is 
desired.  Then  substituting  the  above  values  in  Formula  (N-),  we  have 

20,000  r          4     /  144V] 
P  =  -m-[1+3Mo\2A8)  \  = 

In  this  example,  the  direct  unit  stress  is  20,000  -=-  100  =  200  Ibs.  So 
the  stress  due  to  cross  bending  is  868  -  200  =  668  Ibs.  The  working  unit 


11C  STRUCTURAL  ENGINEERING 

stress  on  timber  should  be  taken  at  about  1,200  Ibs.     So  the  above  column 
will  safely  support  more  than  20,000  Ibs. 

Example  2.  What  direct  unit  stress  would  the  above  column  safely 
sustain  if  the  working  unit  stress  be  taken  as  1,200  Ibs.?  Here  P/A  is 
desired.  Then  using  Formula  (O),  we  have 

P  1,200 

144  i  =        per  sq'  m' 


4  /  144  yi 

+  3,000  \2.88/  J 


Then  the  safe  load,  or  P,  would  be  277  x  100  =  27,700  Ibs. 

What  is  usually  given  is  the  load  and  the  length  of  the  column.  In 
that  case  the  problem  is  to  design  a  column  outright  that  will  safely  carry 
this  known  load.  In  all  such  cases  it  is  practically  a  matter  of  first 
guessing  a  column  which  we  think  will  satisfy  the  conditions  and  then 
modifying  our  guess  until  the  conditions  are  satisfied,  as  will  be  shown  in 
the  following  problems. 

Example  3.  Design  a  timber  column  10  ft.  long  to  carry  a  load  of 
20,000  Ibs.  Taking  1,200  Ibs.  as  the  working  unit  stress,  then  p  will  be 
1,200.  Using  Formula  (N),  we  have 


Here  it  is  seen  that  A  and  r  are  unknown.  First  assume  that  a  column 
9x9  ins.  will  do.  Then  A  =  81  and  r  =  2.6.  Substituting  these  values  in 
the  last  equation,  we  have 


which  is  too  small,  or,  in  other  words,  the  column  is  too  large.     So  next 
try  an  8  x  8-in.  section.     Then  we  have 


which  is  too  large,  that  is,  this  column  is  too  small.  So  the  correct  size  is 
between  the  two.  However,  a  9  x  9-in.  section  would  be  used,  as  more 
than  likely  nothing  between  the  two  could  be  obtained. 

In  case  of  steel  columns,  the  straight  line  Formula  (Q)  will  be  used, 
although  the  Formula  (P)  could  be  used. 

Example  4-  Design  a  steel  column  25  ft.  long  to  support  a  load  of 
160,000  Ibs.  Using  Formula  (Q),  we  have 

p  =  16,000  -70—. 

The  direct  unit  stress  in  accordance  with  economy  should  not  be  less  than 
9,000  Ibs.,  and  it  rarely  ever  exceeds  14,000  Ibs.  So,  guessing  in  this 
case,  as  the  column  is  short,  that  it  will  be  13,000  Ibs.,  and  dividing  the 
load  by  that,  we  have 

160,000 


From   Table    3   it   is    seen   that   2—  [s    12"    x    20.5  #   have    about   this 


THEOEETICAL  TEEATMENT  OF  COLUMNS  HI 

section.  The  r  for  these  two  channels  in  reference  to  the  gravity  axis 
perpendicular  to  their  webs  will  be  the  same  for  the  two  as  it  is  for  one 
channel,  which  is  given  in  the  same  table  as  4.61.  The  radius  in  ref- 
erence to  the  gravity  axis  parallel  to  the  webs  can  always  be  made  equal 
to  the  one  given  in  the  table  by  moving  the  channels  far  enough  apart. 
Now  substituting  the  above  value  of  r  in  the  above  formula,  we  have 


p  =  16,000  -  70      j  =  11,450  Ibs. 

Then  dividing  this  into  the  load,  we  have 

160,000 

'11450    =14'°  sq*  m'  (about)> 

which  is  more  than  the  area  of  the  channels  assumed.  So  try 
2—[s  12"  x  25#  =  14.7n".  Then  r,  from  Table  3,  is  4.43,  and  our 
formula  becomes 


p  =  16,000  -70  -        -11,260  Ibs. 


Then  dividing  this  into  the  load,  we  have 
160,000 


which  agrees  very  closely  with  the  area  of  the  2  —  [s  12"  x  25  #,  which 
would  therefore  be  used.  The  channels  composing  a  column  as  just 
designed  would  be  latticed  together. 

The  method  of  applying  the  column  formula  in  the  last  example  is 
general,  but  the  form  of  column  is  only  one  of  many.  It  is  beyond  the 
limits  of  practicability  to  have  the  radii  of  gyration  tabulated  for  each 
form.  However,  the  approximate  value  of  the  radius  for  any  form  in 
general  use  can  be  obtained  from  Table  10,  where  it  is  given  in  terms  of 
the  height  and  width  of  the  section,  as  shown.  After  the  approximate 
area  of  the  cross-section  is  obtained  by  dividing  the  load  by  an  assumed 
intensity,  as  above,  the  section  can  be  selected  and  then  the  corresponding 
approximate  radius  of  gyration  can  be  obtained  from  Table  10.  In  this 
way  the  assumed  section  can  be  tested,  and,  if  found  to  be  about  correct, 
the  exact  radius  can  be  computed  and  substituted  in  the  column  formula, 
and  the  section  modified  slightly,  if  necessary,  to  satisfy  this  last  require- 
ment. The  form  of  steel  columns  depends  upon  the  kind  of  structure  and 
the  position  they  occupy  in  the  structure  as  well  as  to  the  loads  they 
support. 

The  value  of  p  for  the  different  values  of  L/r  could  be  taken  from 
such  a  diagram  as  shown  in  Fig.  98  if  it  be  drawn  to  a  sufficiently  large 
scale,  or  a  table  giving  the  same  can  be  computed.  The  latter  is  often 
used  in  designing  offices. 

In  case  a  column  is  fixed  at  the  ends,  one-half  of  the  length  of  the 
column  is  taken  as  L,  and  if  fixed  at  one  end  only,  two-thirds  of  the 
length  is  taken  as  L  in  the  column  formulas.  The  reason  for  so  doing  is 
readily  seen,  as  the  actual  length  of  the  column  as  far  as  bending  is 
concerned  in  the  first  case  is  the  distance  between  the  points  of  contra- 
flexure,  which  is  about  one-half  of  the  length  of  the  column;  and  in  the 


112 


STEUCTUEAL    ENGINEEEING 


st-cond  case  it  is  the  distance  from  the  free  end  to  the  point  of  contra- 
flexure,  which  is  about  three-fourths  of  the  length  of  the  column,  the  same 
as  in  the  case  of  beams.  (See  Arts.  68  and  69.) 

75.    Columns    Eccentrically   Loaded. — Whenever 

possible,  the  load  on  any  column  should  be  applied  in  the 
center  of  gravity  of  its  cross-section.  However,  in  some 
cases  this  is  not  possible.  In  such  cases  the  bending 
stresses  due  to  the  eccentric  loading  must  be  considered. 
Let  AB  (Fig.  99)  represent  a  column,  which  bends 
as  indicated,  due  to  the  eccentrically  applied  load  P.  Let 
e  represent  the  eccentric  distance,  A  the  maximum  deflec- 
tion, and  x  and  y  the  co-ordinates  to  the  elastic  curve  re- 
ferred to  A  as  origin.  Now  the  bending  moment  at  any 
point  x  distance  from  A  is  equal  to 

*&•=-*<•+»>• 


For  convenience  let  k  =  yP/EI.     Then  we  have 
d~v 


Fig. 


Now  multiplying  both  sides  of  the  equation  by  2dy,  we  have  the  form 


dx* 
Integrating  (dx  constant),  we  have 


(1) 


Now,  as  is  readily  seen.,  dy/dx  =  0  when  y  =  A.    Therefore,  C,  =  k*(e  +  A)2. 
Then  substituting  this  value  of  C,  in  (1),  we  have 


Now  extracting  the  square  root  and  transposing,  we  have 


Integrating  this  equation,  we  have 


Now  y  =  A  when  x  =  L/2;  therefore,  C,,  =  (L/2  -  ir/2k).     Then  substi- 
tuting this  value  in  (2),  we  have 


Now  y  =  0  when  x  =  0.    So,  substituting  0  in  (2)  for  both  y  and  x,  we 


THEORETICAL  TREATMENT  OF  COLUMNS  113 


e,,=-r 


Then  by  equating  the  two  values  of  C,, ,  we  have 

_l/5in-i     *     \      /£_,r\ 
k\  e  +  &J      \2      2k  J' 

from  which  we  obtain 


cos 


Then  substituting  this  value  of  A  in  (3)  ,  we  have 


Now  transposing  and  reducing,  we  have 

kL 


But,,  according  to  trigonometry, 

kL 


e  +  y        kL 

=  —  COST- 

But,  cos  ir/2  —  0,  and  sin  ?r/2  =  1.     Then  the  last  equation  reduces  tc 

/         kL\     e  +  y       kL 
cos(kx--)=—  cos-. 

But  again,  according  to  trigonometry, 

/,        *L\  kL  ,    kL     e  +  y       kL 

cos  (  kx  -  ——  1  =  cosAr^cos  -^-  +  smfcxsiu  -—  =  —  ^-cos  —  -. 

\  a  I  <>  a          e  fy 

Dividing  through  by  coskL/2,  we  have 

kL     e  +  y 
coskx  +  sinAr^rtan  —  -  =  -  -  . 

But,  tan   kL/2  =   1   -   cosfcL/sin&L.      Then   substituting  this   value   of 
tan  kL/2  in  the  last  equation,  and  reducing,  we  have 

el  sink  (L  -  x  )  +  si 
•    -  — 


.    ir 
sinArL 


(5). 


Now  from  this  last  equation,  y  can  be  computed  for  any  point  along 
the  column,  and  then  the  lever  arm  (e  +  y)  is  known  and  hence  the  stress 
due  to  bending  at  any  point  along  the  column  can  be  determined,  as  we 
have 


J14  STRUCTURAL  ENGINEERING 

according  to  Art.  53.     (d  is  here  the  distance  from  the  neutral  axis  to  the 
outermost  element  in  compression.) 

Then  by  adding  the  unit  compressive  stress  (/)  thus  obtained  to  the 
direct  unit  stress  (P/A),  we  have  the  maximum  unit  stress  at  the  point 
considered.  This,  of  course,  will  be  a  maximum  when  y  =  A.  Now 
y  =  A  when  x  -  L/2.  Then,  substituting  this  value  of  x  in  (5)  and 
reducing,  we  have 

IcT 
e(sec  —  -  1)  =  A  ...................................  (6). 

This  last  equation  (6)  can  be  used  for  determining  A,  the  maximum 
deflection.  After  this  is  determined,  the  maximum  stress  can  readily  be 
determined  as  stated  above.  For  the  maximum  compressive  stress  we 
then  have  the  formula 

P= 


where  p  is  the  maximum  compressive  unit  stress  in  pounds,  7  the  moment 
of  inertia  of  the  cross-section  of  the  column,  A  the  area  of  the  cross- 
section  in  square  inches  (which  is  assumed  to  be  constant),  and  P,  e>  and 
A  are  as  stated  above, 


CHAPTER  VI 


j: 


RIVETS,   PINS,    ROLLERS,   AND    SHAFTING 

RIVETS 

76.  Kinds  of  Stress.  —  Rivets  may  fail  by  shearing  off  transversely, 
by  bearing,  that  is,  by  zrushing  against  the  metal  through  which  they 
pass,  or  by  bending,  as  a  beam.     So,  therefore,  we  have  to  consider  shear- 
ing, bearing,  and  bending  stresses  on  them. 

77.  Shearing  Stress.  —  The  shear  on  a  rivet  at  any  cross-section  is 
equal  to  the  algebraic  sum  of  the  forces  on  either  side  of  the  section  — 
just  the  same  case  as  a  beam.     Let  A  and  B   (Fig.  100)  represent  two 
bars  held  together  by  one  rivet,  as  shown.     Let  P  be  the  tensile  stress  on 
each  bar.     Then  the  bar  B  would  exert  forces  along 

ab  upon  the  rivet,  the  sum  of  which  (resolved  along  the 

bar)  would  be  equal  to  P,  while  the  bar  A  would  exert 

forces  upon  the  rivet  along  cd,  the  sum  of  which  (re- 

solved along  the  bar)  would  be  equal  to  P.     The  forces 

along  ab,  acting  against  the  forces  along  cd,  tend  to 

shear   the    rivet   off   transversely.     This    shear,   as    is 

readily  seen,  varies  from  0  at  each  end  of  the  shank  to 

a  maximum  at  the  cross-section  be.     If  it  were  possible 

to  determine  the  intensity  of  these  forces  at  all  points, 

the  shear  at  any  cross-section  along  the  rivet  could  be 

obtained  by  beginning  at  either  end  of  the  shank  and 

simply  summing  up  the  forces  to  the  section  consid- 

ered.    However,  it  is  not  practical  to  do  so,  neither  is 

it  necessary,  for  it  is  readily  seen  that  the  maximum 

shear  occurs  just  between  the  two  bars  at  cross-section 

be.     This  shear,  as  is  evident,  is  equal  to  P,  and  as  it 

is  the  maximum,  it  is  the  only  shear  we  need  consider. 

The  shear  between  the  pieces  connected  is  what  is  generally  referred  to  as 

the  shear  on  rivets,  and  it  is  what  we  shall  consider  as  the  shear. 

The  unit  shearing  stress,  that  is,  the  stress  per  square  inch  on  a  rivet, 
is  obtained  by  dividing  the  shear  (in  pounds)  by  the  area  (in  square 
inches)  of  the  cross-section  of  the  rivet.  Thus  if  S  be  the  shear  in  pounds 
at  a  given  section  of  a  rivet  having  a  cross-section  of  A  square  inches,  and 
V  the  unit  shearing  stress,  we  have 


8 


(p) 


t  P 

Fig.   100 


The  allowable  shear  on  a  given  rivet  is  what  we  usually  desire.  This  is 
obtained  by  multiplying  the  allowable  intensity  per  square  inch  by  the 
area  of  the  cross-section  of  the  rivet.  We  shall  take  this  allowable 
intensity  as  12,000  pounds  per  square  inch  for  shop  rivets  and  10,000 
pounds  for  field  rivets,  as  specified  by  the  A.  R.  E.  Ass'n  in  their  speci- 

115 


116 


STRUCTURAL  ENGINEERING 


fications  for  railroad  bridges.  Then  if  s  be  the  allowable  shearing  stress, 
we  have  s  =  12,000  x  A  for  shop  rivets,  and  *  =  10,000  x  A  for  field  rivets. 
Substituting  in  these  formulas  we  have  for  the  allowable  shear  on  a  J" 
shop  rivet,  5-12,000x0.6013  =  7,216  Ibs.,  and  for  a  J"  field  rivet,  we 
have  s,=  10,000x0.6013  =  6,013  Ibs.  These  values  for  the  different  size 
rivets  are  given  in  column  3  of  Table  11. 

A  rivet  is  in  single  shear  when  there  is  (so  to  speak)  one  tendency 
to  shear  it  off,  double  shear  when  there  are  two  tendencies,  triple  shear 
when  there  are  three,  and  so  on.  If  two  pieces  be  riveted  together,  as 
indicated  in  the  sketch  at  (a)  (Fig.  101),  the  rivets  will  be  in  single 
shear;  and  if  three  pieces  be  riveted  together  as  indicated  in  the  sketch 
at  (6),  the  rivets  will  be  in  double  shear;  and  if  there  were  four  pieces, 
the  rivets  would  be  in  triple  shear.  If  the  pieces  riveted  together  are 
properly  designed,  rivets  in  double  shear  will  have  twice  the  allowable 


P     b    ^ 


(O)  (b) 

.  Fig.   101 


Fig.   102 


shearing  strength  of  rivets  in  single  shear,  and  three  times  in  the  case  of 
triple  shear,  and  so  on.  For  example,  the  shear  on  the  rivet  shown  at 
(a)  (Fig.  101)  would  be  equal  to  P,  while  in  the  case  shown  at  (6)  it 
would  be  equal  to  'P/2,  providing  e  and  d  have  equal  cross-section. 

78.  Bearing  Stress  on  Rivets. — Let  Fig.  102  represent  a  bar 
pulling  -wdth  a  force  of  P  pounds  against  a  rivet  which  passes  through  it. 
The  force  transmitted  to  the  rivet  thereby  will  be  a  normal  force 
uniformly  distributed  from  b  around  to  a,  as  indicated.  Let  t  be  the 
thickness  of  the  bar,  D  the  diameter  of  the  rivet,  and  let  p  be  the  uniform 
bearing  force  per  square  inch.  Now,  the  bearing  force  on  any  infini- 
tesimal strip  through  the  plate,  as  ids,  where  t  is  the  thickness  of  the 
plate,  would  be  ptds.  Now  this  force,  resolved  along  the  bar,  is  equal  to 
ptds  cos#,  where  6  is  the  angle  the  normal  force  makes  with  that  direc- 
tion, as  indicated.  Then  if  the  normal  forces  on  all  such  strips,  from 
i  around  to  a,  be  resolved  along  the  bar  and  summed  up,  we  would  have 
pfZdscosQ  =  P.  But  2dscos6  =  ab  =  D,  the  diameter  of  the  rivet.  So  we 
have 


ptD  =  P. 


(1). 

To  obtain  the  value  of  P,  such  that  the  rivet  would  not  be  too  highly 
stressed,  we  would  substitute  the  working  value  of  p  in  the  above  formula, 
in  which  case  p  would  be  known  as  the  allowable  unit-bearing  stress  on 
the  rivet  and  P  as  the  allowable  bearing  of  the  rivet  on  the  plate  of 
thickness  t.  For  the  unit  value  we  will  take  24,000  Ibs.  per  square  inch 
for  shop  rivets  and  20,000  Ibs.  for  field  rivets,  as  specified  by  the  A.  R.  E. 
Ass'n  in  their  specifications  for  railroad  bridges. 

Now  suppose  the  rivet  shown  in  Fig.  102  to  be  a  J"  shop  rivet,  and 


RIVETS,  PINS,  ROLLERS,  AND  SHAFTING  117 

suppose  the  plate  to  be  J"  thick.      Then  substituting  in   (1),  we  have 

24,000  x  *-  x  1  =  10,500  Ibs.  =  P, 
A       o 

that  is,  the  allowable  bearing  of  a  j"  rivet  on  a  \"  plate  is  10,500  Ibs. 
If  the  thickness  of  the  plate  be  -J"  instead  of  J",  we  would  have 

i-x^ 
4      8 

In  this  way  the  allowable  bearing  for  rivets  of  different  diameters  on 
different  thicknesses  of  plates  can  be  computed.  These  values  are  given 
in  Table  11. 

79.  Bending  Stress  On  Rivets.  —  If  rivets  are  properly  driven, 
bending  stresses  can  be  ignored,  except  in  the  case  of  loose  fillers  as  is 
illustrated  in  Fig.  103.  Here  the  bars  c,  d,  and  e  exert  double  shear  on 
the  rivet  just  the  same  as  the  case  shown  at  e  f3/ 

(b),   Fig.    101,   and   the  bearing   stresses   are  ""  r~*> 


just  the  same,  but  the  idle   fillers,  /  and  g, 

hold  the  bars  apart  so  that  the  rivet  is  really  Fig.  103 

a  beam  loaded  in  the  center  and  supported  at 

the  ends.     Let  k  be  the  distance  from  the  center  of  the  bar  c  to  the  center 

of  the  bar  d,  and  h  the  distance  from  the  center  of  the  bar  c  to  the  center 

of  the  bar  e.     Then  the  maximum  bending  moment  on  the  rivet  is  (P/2)k 

or    (P/2)h.     Now,   according  to    (D),   Art.    53,   the   maximum   bending 

stress  on  the  rivet  per  square  inch  is 


My 

'  "    I 


64 


where  D  is  the  diameter  of  the  rivet. 

The  bending  stress  on  a  rivet  should  always  be  combined  with  the 
shearing  stress  according  to  Art.  62  so  as  to  obtain  the  maximum  stress. 

Details  should  be  so  contrived  as  to  avoid  bending  on  rivets.  In 
fact,  there  is  practically  no  excuse  for  placing  rivets  so  that  they  are 
subjected  to  bending  of  any  consequence. 

80.  Examples  in  Determining  Number  of  Rivets.  —  Suppose 
that  each  of  the  bars  a  and  b,  shown  at  (o),  Fig.  101,  be  J"  thick,  and 
suppose  the  stress  P  to  be  50,000  Ibs.,  how  many  }"  shop  rivets  would  be 
required  to  transmit  this  stress? 

The  allowable  single  shearing  stress  on  one  f  "  shop  rivet,  from  Art. 
77,  is  12,000  x  7r(|)2  -r  4  =  12,000  x  0.4418  =  5,301  Ibs.  Then  the  number 
of  rivets  required  for  shear  is  50,000  ^  5,301  =  9.4  rivets  (use  10). 

The  allowable  bearing  stress  on  one  f"  shop  rivet,  from  (1),  Art.  78, 
is  24,000  x  J  x  f  =  4,500  Ibs.  Then  the  number  required  for  bearing  is 
50,000  4-  4,500  =  11  rivets  (about).  So  11  rivets  is  the  number  required. 

Now  suppose  the  bars  to  be  f  "  thick  instead  of  J".  The  allowable 
shearing  stress  on  each  rivet  would  not  change.  So  10  rivets  would  be 
required  for  shear  the  same  as  before.  But  the  allowable  bearing  stress 
is  now  24,000  x  J  x  }  =  6,750  Ibs.  Then  the  number  of  rivets  required 


STRUCTURAL  ENGINEERING 

for  bearing  is  50,000  -f  6,750  =  7.4  (use  8).  So  in  that  case  10  rivets,  as 
required  for  shear,  would  be  used,  as  it  is  the  greater  number. 

As  another  example,  suppose  the  bar  c,  shown  at  (6),  Fig.  101,  to  be 
y  thick,  and  each  of  the  bars  d  and  e  to  be  §"  thick,  and  suppose  the 
stress  P  to  be  60,000  Ibs. ;  how  many  f "  shop  rivets  would  be  required  to 
transmit  this  stress?  The  allowable  double-shearing  stress  on  one  f" 
shop  rivet  is  2  x  12,000  x  0.3068  =  7,364  Ibs.  Then  the  number  of  rivets 
required  for  shear  is  60,000  -^  7,364  =  8.15  rivets  (use  9).  The  bearing 
here,  in  one  direction,  is  on  a  \"  bar,  while  in  the  other  direction  it  is  on 
two  f "  bars,  which  is  equivalent  to  one  f  "  bar.  So  the  bearing  stress  on 
the  \"  bar  (c)  will  be  the  greater.  The  allowable  bearing  stress  on  one 
rivet  is  24,000  x  -J  x  f  =  7,500  Ibs.  So  the  number  of  rivets  required  for 
bearing  is  60,000  -j-  7,500  —  8  rivets.  Here  the  number  of  rivets  used 
would  be  9,  the  number  required  for  shear. 

The  allowable  bearing  and  shearing  intensities  for  rivets  are  given 
in  Table  11.  However,  the  student  should  know  how  to  compute  these 
intensities. 

PINS 

81.  Kinds  of  Stress. — The  same  kind  of  stresses  occur  on  pins  as 
on  rivets.     However,  in  the  case  of  pins,  the  bending  stresses  are,  as  a 
rule,  the  most  serious. 

82.  Shearing  and  Bearing'  Stresses. — The  shearing  and  bearing 
stresses  on  pins  are  determined  exactly  as  on  rivets.     The  allowable  unit 
intensities  for  shear  and  bearing  are  12,000  and  24,000  Ibs.,  respectively 
— the  same  as  for  shop  rivets.     The  shearing  stress  very  rarely  affects 
the  design  of  a  pin,  while  the  same  is  practically  true  of  the  bearing  stress. 
However,  the  bearing  affects  the  details  of  the  other  parts  of  a  structure 
as  each  member  connected  must  have  the  proper  thickness  of  bearing  on 
the  pin.    As  an  example,  what  would  be  the  required  thickness  of  bearing 
to  transmit  a  stress  of  250,000  Ibs.  to  a  6-in.  pin?     The  allowable  bearing 
of  a  plate  1  in.  thick  on  this  pin,  according  to  Art.  78,  is  24,000  x  1  x  6  = 
144,000   Ibs.      Then   the  thickness   of  the   bearing  required   is   250,000 
-f  144,000  =  1.76  ins. 

83.  Bending  Stresses  on  Pins. — The  forces  applied  to  pins  are 
assumed  to  be  applied  at  the  center  of  bearings  of  the  pieces  connected. 

The  bending  moments  on  pins  can  be  computed  most 
readily  by  utilizing  the  proposition  in  Art.  66.  For  ex- 
ample, let  AB  (Fig.  104)  represent  a  pin  acted  upon  by 
forces  Fa,  F^,  etc.,  applied  at  sections  a,  b,  c,  etc.,  as 
shown,  and  let  Ma,  Mb,  etc.,  and  Sa,  Si>,  etc.,  represent 
the  bending  moments  and  shears,  respectively,  at  the  sec- 
tions a,  b,  c,  etc.  It  is  obvious  that  the  bending  moments 
at  sections  a  and  g  are  equal  to  zero.  Then,  by  starting 
at  either  of  these  sections  we  can  determine  the  bending 
moment  at  each  of  the  other  sections,  according  to 
Art.  66. 
So,  starting  from  a,  the  bending  moment  at  6  is 

=  (a6)F0; 


RIVETS,  PINS,  EOLLERS,  AND  SHAFTING  119 

at  c  it  is 

Mc 
at  d  it  is 

Md 
at  e  it  is 

Me 
at  /  it  is 


Mf  =  Mc  +  (ef)Se  =  Me  +  ef(Fa  -Fb  +  Fc-  Fd 


In  determining  the  bending  moments  in  this  manner,  the  maximum 
is  readily  observed.  Care  should  be  taken  in  regard  to  algebraic  signs. 
The  shear  in  some  cases  will  be  minus,  as  is  readily  seen,  and  consequently 
the  moments  in  some  cases  may  be  minus. 

As  a  numerical  example,  let  the  lever  arms  be  as  follows  : 

ab  =  2  in.,  de  =  l  in., 

6c  =  lJ  in.,  ef=ll>  in., 

cd  =  1  in.,  /<7  —  ^  in-  ; 

and  let 

Fa=  150,000  Ibs.,  Fe  =  125,000  Ibs., 

F6  --200,000  Ibs.,  Ff  =  -200,000  Ibs., 

Fc  -125,000  Ibs.,  F,  =  150,000  Ibs.; 
Fd  =  -150,000  Ibs., 
then 

Sa  =  150,000  Ibs.,  Sd  =  -75,000  Ibs., 

£b  =  -50,000  Ibs.,  Se  =  50,000  Ibs., 

Se  =  75,000  Ibs.,  5;  =  -150,000  Ibs. 

Now,  by  substituting  these  values  in  the  above  equations,  we  have 

M6  =  2  x  150,000  =300,000  in.  Ibs., 

Mc  =  300,000-  75,000  =  225,000  in.  Ibs., 
Md  =  225,000+  75,000  =  300,000  in.  Ibs., 
Me  =  300,000-  75,000  =  225,000  in.  Ibs., 
Mf  =  225,000+  75,000  =  300,000  in.  Ibs., 
Mg  =  300,000  -  300,000  =  0. 

Here  it  is  seen  that  the  maximum  moment  on  the  pin  would  be 
300,000  inch  pounds,  which  happens  to  occur  at  three  points  —  b,  d,  and  /. 

After,  the  maximum  bending  moment  is  thus  obtained,  the  bending 
stress  is  computed  according  to  (D),  Art.  53,  just  the  same  as  if  the  pin 
were  a  beam.  Thus,  in  the  last  example,  suppose  the  pin  to  be  6"  in 
diameter.  Then  we  would  have 


/  =  300,000x3  +  7r      =  14,150  Ibs.  (about). 

The    allowable   bending   stress    intensity    for   pins,   as    specified   by   the 
A.  R.  E.  Ass'n,  is  25,000  Ibs.  per  square  inch. 

Let  D  be  the  diameter  of  any  pin,  M,  the  bending  moment,  and  we 
have 


120  STRUCTURAL  ENGINEERING 


2  T   64 
Transposing,  we  have 


(i) 


Now  by  substituting  25,000  Ibs.  for  /  in  this  last  equation,  the  diameter 
of  pin  required  for  any  given  bending  moment  can  be  determined.  The 
diameters  required  for  given  moments  are  given  in  Table  12.  Here  the 
moments  are  given  in  inch  pounds,  and  /  taken  as  25,000  pounds. 

Often  the  members   bearing   on   a  pin   act  upon  it   from   different 
directions,   as   shown   in    Fig.    105.      In   such   cases   the   stresses   in   the 
members  are  resolved  horizontally  and  vertically  and  the  bending  moments 
determined  in  each  plane,  and  the  maximum  or  resultant  moment  is  then 
obtained  by  extracting  the  square  root  of  the  sum 
of   the   squares   of   these   maximum    horizontal   and 
vertical  moments.     Thus,  in  the  case  shown  in  Fig. 
F2     105,   F3   would   be   resolved   horizontally   and   ver- 
Fig.  105  tically.     Then  the   horizontal  component  would  be 

included  with  the  forces  Fl  and  F2  in  the  computa- 

tions for  the  horizontal  moments,  and  the  vertical  component  would  be 
included  with  the  force  F4  in  the  computations  for  the  vertical  moments. 

Let  Mh  represent  the  maximum  horizontal  moment,  and  let  Mv 
represent  the  maximum  vertical  moment;  then  the  maximum  or  resultant 
moment  would  be 


for  which  the  pin  would  be  designed. 

ROLLERS 

84.  Allowable  Pressure.  —  In  modern  practice,  the  allowable  pres- 
sure on  rollers  is  obtained  by  the  use  of  an  empirical  formula,  which  has 
the  form 

p  =  cD  ...............  ..............................  (1), 

where  p  =  the  allowable  pressure  per  lineal  inch  of  roller  and  D  =  the 
diameter  of  the  roller  in  inches,  and  c  =  a  constant. 

Prof.  A.  Marston*  found  that  for  the  elastic  limit  of  soft  steel  rollers 
c  was  about  880.  (See  paper  Trans.  A.  S.  C.  E.,  Vol.  32.)  For  the 
allowable  pressure,  c  should  be  taken  as  about  one-half  of  this.  Then  for 
the  allowable  pressure  per  lineal  inch  on  soft  steel  rollers,  we  have 

p  =  440  D  ..........................................  (2). 

The  formula,  for  medium  steel  rollers  as  specified  by  the  A.  R.  E. 
Ass'n,  is 

p  =  600  D  ..........................................  (3). 

This  is  the  formula  now  in  general  use,  as  rollers  are,  as  a  rule,  made 
of  medium  steel. 


*Dean,  Engineering  Division,  Iowa  State  College. 


RIVETS,  PINS,  ROLLERS,  AND  SHAFTING  121 

The  formula  is  very  readily  applied.  As  a  practical  example,  sup- 
pose a  load  of  600,000  pounds  be  supported  upon  six  6"  rollers;  what  will 
be  the  required  length  of  each? 

Substituting  in  Formula  (3),  we  have  p  =  600  x  6  =  3,600  Ibs.  Then 
the  total  linear  inches  required  is  600,000  -=-  3,600  =  166.6  ins.;  and  as 
there  are  six  rollers,  the  length  of  each  will  be  166.6  •*•  6  =  27.7  ins. 

Rollers  should  always  be  as  large  in  diameter  as  is  consistent  with 
accompanying  details. 

SHAFTING 

85.  Stress  Due  to  Torsion. — Shafting  used  to  transmit  power  for 
operating  movable  bridges,  as  well  as  in  all  other  machinery,  is  subjected 
to  a  twisting  about  the  longitudinal  axis  known  as  torsion.  The  stresses 
resulting  therefrom  are  known  as  torsion  stresses.  This  stress  is  the  same 
as  the  shearing  stress  in  a  beam,  except  the  action  on  each  particle  is 
perpendicular  to  a  radius  through  the  center  of  rotation,  and  the  stress 
varies  directly  as  the  distance  out  from  this  center  of  rotation.  The 
difference  results  owing  to  the  stress  being  caused  by  a  tendency  of 
rotation  instead  of  a  tendency  of  translation,  as  in  the  case  of  a  beam. 

Let  AB  (Fig.  106)  represent  a  round  steel  shaft  subjected  to  torsion. 
Suppose  this  shaft  be  cut  off  at  ab  and  spliced  by  means  of  inserted  steel 
pins  which  fit  tightly  into  drilled  holes. 
Now  it  is  readily  perceived  that  the  tor- 
sion (twisting  of  the  shaft)  tends  merely 
to  shear  each  of  these  pins  off  transversely 
and  perpendicularly  to  a  radius  through 
the  center  of  rotation,  and  as  far  as  the  Fig.  IOG 

torsion  is  concerned,  the  pins  have  only  a 

direct  shearing  stress  on  them.  Now,  evidently,  the  material  particles  in 
every  cross-section  of  the  shaft  are  subjected  to  exactly  the  same  kind  of 
stress  from  torsion  as  the  pins,  as  the  action  causing  the  stress  is  the  same. 

This  shearing  stress  varies  directly  as  the  distance  out  from  the 
center  of  the  shaft.  This  is  readily  seen  by  considering  again  the  pins  at 
section  ab.  Each  pin  will  be  distorted  a  certain  amount,  and  as  the 
tendency  of  rotation  is  about  the  center  of  the  shaft,  it  is  evident  that  the 
distortion  of  the  pins  will  vary  directly  as  their  distance  out  from  the 
center  of  the  shaft,  and  hence  the  stress  on  them  will  so  vary.  Then, 
evidently,  if  the  stress  on  these  pins  varies  directly  as  their  distance  out 
from  the  center  of  the  shaft,  the  stress  on  the  material  particles  at  every 
cross-section  of  the  shaft  will  vary  directly  as  their  distance  out  from  the 
center  of  the  shaft. 

So  we  have  thus  far  shown  that  the  stress  on  the  shaft  due  to  torsion  is 
a  transverse  shearing  stress  acting  perpendicularly  to  the  radii  of  rotation, 
and  that  it  varies  directly  as  the  distance  out  from  the  center  of  the  shaft. 
Now  it  remains  yet  to  determine  the  intensity  of  the  stress.  The  stress, 
of  course,  will  depend  upon  the  torsion,  which  is  directly  proportional  to 
the  moment  about  the  center  of  the  shaft  of  the  force  producing  the 
torsion.  Let  P  represent  the  force  producing  the  torsion  of  the  shaft  AB 
(Fig.  106)  and  let  d  be  its  lever  arm.  Then  the  moment  of  this  force, 
which  produces  the  torsion,  is  Pd.  This  would  be  known  as  the  "torque." 
Now  it  is  evident  that  this  moment  Pd,  or  torque,  must  be  balanced  at 


122  STRUCTURAL  ENGINEERING 

each  cross-section  of  the  shaft  by  the  moments  of  the  torsion  stresses,  at 
each  section,  about  the  center  of  the  shaft. 

Let  Fig.   107   represent  an  enlarged  cross-section  of  the  shaft  AB 
(Fig.  106).    As  the  torsion  stresses  vary  directly  as  the  distance  out  from 
the  center  of  the  shaft,  the  maximum  stress  will  evidently  occur  at  the 
outermost  elements.     Let  S  be  this  stress  per  square  inch, 
and  let  r  be  the  radius  of  the  shaft.     Then  the  stress  per 
square  inch  out  unit  distance  will  be  S/r,  and  the  stress  per 
square  inch  out  any  distance  z  will  be  (S/r)z.     But  as  the 
stress  varies  continuously  outward   from  the  center  of  the 
shaft  it  is  evident  that  there  will  not  be  a  square  inch  of 
Fig.  107        material  out  z  distance  having  a  stress  of  (S/r)z,  but  only 
an  infinitesimal  area  da  of  material  having  that  stress.     So 
the  actual  stress  or  force  out  any  distance  z  is  (S/r)zda,  and,  of  course, 
the  moment  of  this  about  the  center  of  the  shaft  is  (S/r)z2da.     Now  it  is 
evident  that  the  stress  is  the  same  on  all  material  equal  distances  out  from 
the  center  of  the  shaft,  so  da  can  be  taken  as  a  circular  strip  of  width  dz 
and  length  2irz.    Then  we  have  da  =  2-rrzdz.    Then  the  moment  of  the  tor- 
sion stress  out  z  distance  from  the  center  of  the  shaft  is 

—z2da  =  27T—  z3dz, 


and  summing  up  this  for  the  entire  cross-section,  we  have 


[l 

Jo 


(1) 


Integrating,  we  have  Pd  =  S7rrz/2,  and  transposing, 
2Pd 


from  which  the  maximum  torsion  stress  per  square  inch  on  any  round 
solid  shaft,  as  the  above,  can  be  computed.  S  =  stress  per  square  inch; 
Pd  =  torque  ;  and  r  =  radius  of  the  shaft. 

In  the  case  of  a  hollow  shaft,  the  integration  of  (1)  would  be  between 
the  limits  of  the  internal  and  external  radii.  Thus,  let  r  be  the  external 
radius  and  r,  the  internal  radius  of  a  hollow  shaft.  Then  we  would  have 


T"»     7  £\  I  Q     7 

pd-2TT—  I    z*dz. 

Integrating,  we  have 

Pd  =  ~ 
Transposing,  we  have 

(3), 


Kt) 


RIVETS,  PINS,  ROLLERS,  AND  SHAFTING  123 

from  which  the  maximum  torsion  stress  per  square  inch  on  any  hollow 
shaft  can  be  computed.  $  =  stress  per  square  inch  ;  r  =  external  radius  ; 
r,  -  internal  radius  ;  and  Pd  the  torque. 

In  case  of  a  shaft  having  a  square  cross-section,  da 
would  be  taken  as  any  infinitesimal  rectangular  area. 
For  instance,  let  Fig.  108  represent  the  cross-section  of  a 
square  shaft,  the  center  of  which  is  at  0.  Then,  for  the 
moment  of  the  torsion  stress  on  an  infinitesimal  area  da 
out  any  distance  z  from  0,  we  have  (S/e)z2da.  Here  *V 
is  the  stress  on  the  outermost  element,  which  is  distance  Fig  108 

e  from  0.     Let  Pd  be  the  torque,  and  we  have 


e 
Expressing  z  in  terms  of  the  rectangular  co-ordinates  x  and  y,  we  have 

Pd  =  -  (2x2da  +  2y2da). 

But  (^x~da  +  Hy2da)  is  the  polar  moment  of  inertia  of  the  cross-section, 
which  is  designated  as  J  in  Art.  49.     Then  we  have 


Transposing,  we  have 

S=l™>£  ..........................................  (4), 

J 

from  which  the  maximum  torsion  stress,  not  only  on  square  shafting  but  on 
any  shaft  whatever,  can  be  computed.  S  -  stress  per  square  inch;  Pd  = 
torque  ;  J  =  polar  moment  of  inertia  ;  and  e  =  the  distance  to  the  outermost 
element.  Formula  (4)  is  a  general  formula,  as  it  applies  to  any  form  of 
cross-section. 


CHAPTER  VII 

MAXIMUM  REACTIONS,  SHEARS,  AND  BENDING  MOMENTS 

ON  SIMPLE  BEAMS  AND  TRUSSES,  AND 

STRESSES  IN  TRUSSES 

86.  Maximum  Reactions  on  Simple  Beams. — The  reactions  on 
simple  beams  in  all  cases  can  be  determined  by  taking  moments  about  the 
supports  as  explained  in  Art.  54.     In  the  case  of  uniform  dead  load,  the 
two  reactions  are  equal  and  each  is  known  directly  as  being  equal  to  half 
of  the  dead  load  supported.      In  case  the  dead  load  is  not  uniformly 
distributed,   the   reactions    are   obtained   by   taking   moments   about   the 
supports,  as  stated  above.     In  all  cases  of  dead  load,  the  reactions  are 
fixed  in  intensity,  as  the  loads  are  fixed  in  position,  but  for  live  load  the 
case  is  different  as  the  reactions  vary  with  the  position  of  the  loads,  and 
the  maximum  reaction  occurs  when  the  loads  are  in  one  certain  position. 
If  the  live  load  be  a  uniform  load,  the  reactions  at  the  two  supports  will 
be  a  maximum  at  the  same  time  and  that  will  be  when  the  load  extends 
over  the  full  length  of  the  beam,  in  which  case  each  reaction  is  known 
directly  as  being  equal  to  half  of  the  load  supported.     The  live  load  often 
consists  of  wheel  loads,  as  in  the  case  of  locomotives,  trains,  wagons, 
traction  engines,  etc.,  where  the  loads  or  wheels  are  a  fixed  distance  apart. 
The  maximum  reaction  due  to  such  loading  will  occur,  as  is  readily  seen, 
when  the  beam  is   fully  loaded  and  the  heaviest  loads  are  as  near  as 
possible  to  the  support  considered. 

87.  Maximum  Shear  on  Simple  Beams.— The  reactions  on  simple 

beams,  in   all  cases,  can  be  determined  by   taking   moments   about  the 

L 


Fig.   109 


supports,  as  stated  in  the  preceding  article,  and  after  the  reactions  are 
known  the  shear  at  any  vertical  section  of  a  beam  is  determined  by  simplv 
adding  up  the  forces  on  either  side  of  the  section,  beginning  at  the  end  of 
the  beam,  as  explained  in  Art.  52.  This  much  applies  in  all  cases  whether 

124 


MAXIMUM  REACTIONS,  ETC.  125 

the  shear  be  a  maximum  or  not.  The  maximum  shear  on  any  vertical 
section  of  a  beam,  due  to  dead  load,  is  readily  obtained  as  such  loads  are 
fixed  in  position  and  hence  the  shear  obtained  by  adding  up  the  forces 
on  either  side  of  any  section  is  the  maximum  for  that  section,  but  in  the 
case  of  live  load  the  maximum  shear  is  not  so  readily  obtained  for  the 
shear  on  every  vertical  section  of  a  beam  changes  continually  as  such 
loads  move  over  the  beam,  and  hence  the  maximum  shear  will  occur  when 
the  load  is  in  one  certain  position.  As  an  example  of  live  load,  let  AB 
(Fig.  109)  represent  a  simple  beam  supporting  a  single  moving  load 
P,  and  let  R  and  Rl  represent  the  reactions  at  A  and  B,  respectively,  due 
to  this  load  when  at  any  point  on  the  beam. 

The  shear  on  any  short  strip  through  the  beam,  as  abed,  due  to  P, 
can  be  expressed  as 


when  the  load  is  at  any  point  to  the  right  of  the  strip.  The  shearing 
force  exerted  upon  the  strip  will  then  act  as  indicated.  It  is  readily  seen 
that  this  shear  on  the  strip  varies  directly  as  x  and  hence  will  be  a 
maximum  when  x  —  m.  Then  if  we  lay  off  the  vertical  line  gh  equal  (by 
scale)  to  Pm/L  and  draw  the  line  hk,  any  ordinate  to  this  line  hk  graph- 
ically represents  the  shear  on  the  strip  abed  when  the  load  is  directly  over 
the  ordinate.  For  example,  the  ordinate  y  represents  the  shear  on  the 
strip  when  the  load  is  in  the  position  shown,  and  any  other  ordinate  as  2 
represents  the  shear  on  the  strip  when  the  load  is  over  that  ordinate. 
When  the  load  is  at  any  point  to  the  left  of  the  strip,  the  shear  on 
the  strip  can  be  expressed  as 


but  the  shearing  forces  exerted  upon  the  strip  will  then  act  in  the  oppo- 
site direction  to  those  shown.  This  change  in  the  direction  of  action  of 
the  shearing  forces  takes  place  the  instant  the  load  passes  the  strip.  We 
would  say  that  the  shear  changes  signs  as  the  load  passes  the  strip  and 
that  the  shear  on  the  strip  is  positive  when  the  load  is  on  one  side,  and 
negative  when  on  the  other. 

It  is  seen,  from  the  last  equation,  that  AS"  will  be  a  maximum  when 
R  has  its  least  value.  This,  as  is  seen,  occurs  when  the  load  is  just  to 
the  left  of  the  strip,  that  is,  when  x  =  m  +  dx,  but,  as  dx  can  be  taken  as 
an  infinitesimal,  we  would  say  when  x  —  m.  Then  if  we  lay  off  the  vertical 
line  fg  equal  to  (Pm/L)  -P=(R-P)  and  draw  the  line  fe  we  have  the 
shear  on  the  strip  graphically  represented  for  the  load  P  at  any  point  to 
the  left  of  the  strip,  as  any  ordinate  r,  to  the  line  fe,  represents  the  shear 
on  the  strip  when  the  load  is  over  that  ordinate.  Then  we  have  the  variation 
of  the  shear  on  the  strip  abed  due  to  the  load  P,  as  it  moves  over  the  beam 
from  B  to  A,  fully  represented  by  the  ordinates  to  the  broken  line  efhk. 
It  is  seen  from  this  that  the  loads  on  the  two  sides  of  any  vertical  section 
of  a  beam  really  neutralize  each  other  as  regards  the  shear  they  produce 
on  the  section,  and  hence  it  appears  that  the  maximum  shear  would  occur 
when  the  loads  are  on  one  side  only  and  extending  on  that  side  from  the 


126 


STRUCTURAL  ENGINEERING 


end  of  the  beam  up  to  the  section.  This  is  absolutely  true  in  the  case  of  a 
uniform  live  load,  as  shown  in  Art.  55,  and  is  practically  true  for  prac- 
tically all  classes  of  loading,  yet  it  is  not  absolutely  true  in  some  cases  of 
concentrated  live  loads,  for  sometimes  the  maximum  shear  on  a  vertical 
section  of  a  beam,  due  to  such  loads,  occurs  when  the  loading  extends 
from  one  end  of  the  beam  to  a  little  beyond  the  section  considered.  The 
most  common  instance  of  this  is  that  of  the  typical  system  of  locomotive 
wheel  loads.  In  such  cases  the  exact  position  required  to  produce  maxi- 
mum shear  must  be  determined  by  trial. 


'*' 


As  a  general  case  of  uniform  load, 
let  it  be  required  to  determine  the  maxi- 
mum positive  and  negative  shear  on  a 
beam  12  ft.  long  at  a  point  4  ft.  from 
the  end  due  to  a  uniform  dead  load  of 
Q  100  Ibs.  per  foot  and  a  uniform  live  load 
of  1,500  Ibs.  per  foot.     Let  AB   (Fig. 
Rl  110)    represent  the   beam,   and   let   ba 
represent  the  section  at  which  the  shear 
is  required. 

As  the  dead  load  is  symmetrically 
located  in  reference  to  the  center  of  the 
beam,  the  two  reactions  due  to  dead 
load  are  equal  to  each  other,  and  each 
is  equal  to  6x100  =  600  Ibs.  Then  for 
the  dead-load  shear  at  the  section  ba,  adding  up  from  A  and  taking  R 
(=  600)  as  positive,  we  have 

600  -(4x100)=  +  200  Ibs. 

When  the  uniform  live  load  extends  from  the  end  B  up  to  the  section,  as 
shown,  one  of  the  maximum  live-load  shears  will  occur.  Taking  moments 
about  B  when  the  load  is  in  this  position,  we  have 


(O) 


Fig.   110 


1,500x8x4 
12 


=  +  4,000  Ibs. 


for  the  live-load  reaction  at  A.  Then  for  the  shear  due  to  this  load  at  the 
section  ba,  adding  up  from  A,  we  have  +4,000  Ibs.  as  there  is  no  inter- 
vening live  load.  This  is  the  maximum  positive  shear  at  the  section  ba 
due  to  the  live  load.  It  will  be  readily  seen  that  the  dead  load  and  the 
live  load,  where  the  live  load  extends  from  B  to  the  section  ba,  both  exert 
shearing  forces  at  the  section  which  act  in  the  direction  shown  at  (a), 
hence  the  two  will  add,  and  we  have  200 +  4,000  =  +  4,200  Ibs.  for  the 
maximum  positive  shear  at  the  section  ba,  due  to  the  dead  and  live  load 
combined.  Now  if  the  live  load  extends  from  A  to  the  section  ba  instead 
of  from  B,  the  reaction  at  A,  due  to  the  live  load  in  that  position,  would  be 


R  = 


1,500x4x10 
12 


=  5,000  Ibs. 


Then  for  the  shear  at  the  section  ba  due  to  this  load,  adding  up  from  A, 
we  have  5,000- 6,000  =  - 1,000  Ibs.  which  is  the  maximum  negative  shear 
at  the  section  due  to  the  live  load. 


MAXIMUM  REACTIONS,  ETC.  127 

It  will  be  readily  seen  that  the  live  load  when  extending  from  A  to 
the  section  ba  will  exert  shearing  forces  at  the  section  which  will  act  as 
shown  at  (6),  and  as  this  direction  is  opposite  to  the  direction  of  action 
of  the  shearing  forces  exerted  by  the  dead  load,  the  shear  at  the  section, 
due  to  dead  and  live  load  combined,  will  undoubtedly  be  equal  to  the 
difference  of  the  two  shears.  So  we  have  200  -1,000  =-800  Ibs.  for  the 
maximum  negative  shear  at  the  section  ba  due  to  dead  and  live  load 
combined. 

Now  it  is  seen  that  in  the  above  example  the  maximum  positive  and 
negative  shear  at  the  section  ba,  due  to  dead  and  live  load  combined,  is 
simply  the  algebraic  summation  of  the  two  simultaneous  shears  in  each 
case.  This  will  hold  in  all  cases  provided  the  adding  up  of  the  forces  be 
in  reference  to  the  same  end  of  the  beam  in  each  case.  As  an  illustration, 
let  us  add  up  the  forces  in  the  above  example  from  the  end  B  instead  of 
from  A,  as  we  did  above.  The  reaction  at  B,  due  to  the  dead  load,  is 
600  Ibs.  Then  for  the  dead-load  shear  at  the  section  ab,  adding  up  from 
B,  we  have,  600-  8  x  100  =  -200  Ibs. 

For  the  live-load  reaction  at  B,  when  the  live  load  extends  from  B  to 
the  section  ba,  we  have 


12 

Then  for  the  shear  at  the  section,  adding  up  from  B,  we  have  8,000- 
12,000  =  -4,000  Ibs.  Adding  this  to  the  dead-load  shear,  we  have 
-200  -4,000  =  -4,200  Ibs.,  which  is  the  same  as  we  obtained  by  adding 
up  from  A,  except  the  sign  is  negative  instead  of  positive.  If  the  live 
load  extends  from  A  to  the  section  ba  instead  of  from  B  to  the  section, 
we  have 

1,.500X4X2 


for  the  reaction  at  B  due  to  the  live  load.  Then  for  the  shear  at  the 
section  ba,  adding  up  from  B,  we  have  +  1,000  Ibs.  Now  adding  this  to 
the  dead  load,  as  just  determined  by  adding  up  the  forces  from  B,  we 
have  -200  +  1,000  =  +  800  Ibs.,  which  is  the  same  as  we  obtained  above 
by  adding  up  from  A,  except  that  the  sign  is  positive  instead  of  negative. 
Thus  we  see  that  by  adding  up  the  forces  from  one  end  we  obtain 
positive  shear  and  by  adding  up  the  forces  from  the  other  end  we  obtain 
negative  shear.  There  is  no  mystery  about  this  at  all,  as  is  readily 
perceived  by  referring  to  Fig.  Ill  where  AB  represents  a  simple  beam 
supporting  a  number  of  loads,  and  R  and  Rl  represent  the  reaction  at 
A  and  B,  respectively,  due  to  the  loads,  and  abed  represents  a  very  short 
strip  through  the  beam.  If  we  take  the  reactions  as  positive,  then  any 
force  acting  upward  will  be  positive.  By  adding  up  the  forces  from  A  to 
the  strip  abed  we  really  obtain  the  shearing  force  acting  upon  the  strip 
along  ab  and  by  adding  up  the  forces  from  B  to  the  strip  we  really  obtaiff 
the  shearing  force  acting  upon  the  strip  along  dc.  As  these  two  shearing 
forces  on  any  vertical  section  of  a  beam  are  always  equal  and  opposite,  it 
is  evident  that  the  sum  obtained  by  adding  up  the  forces  from  one  end 
will  be  positive,  while  the  equal  sum  obtained  by  adding  up  the  forces 
from  the  other  end  will  be  negative. 


STRUCTURAL  ENGINEERING 


There  will  be  no  confusion  as  regards  the  sign  of  the  shear  on  any 
se/'tion  of  a  beam  in  any  case,  providing  the  adding  up  of  the  forces,  that 
is.  the  algebraic  summation,  be  started  from  the  same  support. 


i        lad    I          I        1 


R 


Rl 


B 


Fig.   Ill 


88.  Maximum  Bending  Moments  on  Simple  Beams. — The  bend- 
ing moment  at  any  cross-section  of  a  beam,  in  all  cases,  is  equal  to  the 
algebraic  sum  of  the  moments  about  the  section,  of  the  forces  on  either 
side  of  the  section.  In  case  the  load  be  uniformly  distributed  over  the 
entire  length  of  the  beam,  the  maximum  moment  will  occur  at  the  center 
of  the  span,  and  will  be  equal  to  wL2/8  as  shown  in  Art.  56.  This  is  true 
for  all  uniformly  distributed  loads,  either  live  or  dead.  In  the  case  of 
fixed  concentrated  loads,  the  maximum  moment  will  occur  under  one  of 
the  loads  which  can  be  ascertained  readily  by  trial.  However,  it  can  be 
determined  otherwise  as  shown  later. 

If  the  live  load  be  wheel  loads,  it  is  always  necessary  to  first  deter- 
mine the  position  of  the  loads  on  the  beam  for  maximum  moment.  As 
wheel  loads  roll  over  a  beam  it  is  evident  that  the  moment  at  every  section 
is  continually  changing,  and  it  is  the  absolute  maximum  of  these  various 
moments  produced  that  we  desire,  although  it  may  last  only  for  an  instant 
during  the  passage  of  the  load. 

Let  AB  (Fig.  112)  represent 
a  simple  beam  supporting  a  system 
of  wheel  loads.  According  to  Art. 
56,  the  maximum  moment  will  occur 
under  a  wheel.  Let  this  wheel  be 
the  one  marked  C.  Let  W  be  the 
weight  of  all  the  wheels  on  the 
beam,  and  x  the  distance  from  the 
right  support  to  their  center  of 
gravity.  Further,  let  P  be  the 

weight  of  the  wheels  to  the  left  of  wheel  C,  and  b  the  distance  of  their 
center  of  gravity  from  wheel  C,  and  let  2  be  the  distance  from  the  left 
support  to  wheel  C.  Then  the  bending  moment  under  wheel  C  can  be 
expressed  as 


Fig.   112 


Now  from  Fig.   112  it  is  seen  that  x  =  L-a-z.     Substituting  this 
v&me  of  ,r  in  (1),  we  have 


za-^)-Pb (2). 

ZJ 

Now  any  slight  movement  of  tlie  loads  either  to  the  right  or  left  will 
cause  a  small  change  in  2  and  M.     Let  dz  and  dM  represent  this  incre- 


MAXIMUM  REACTIONS,  ETC.  129 

ment  of  z  and  M,  respectively,  due  to  a  slight  movement  of  the  loads. 
Adding  the  increments  in  equation  (2),  we  have 

W 

~[ 
Li 

and  reducing  we  have 

W  W 

M  +  dM  =  ~  [^L  -  az  -  z2]  +  -^  [  dzL  -  adz  -  2zdz]  -  Pb. 
Li  Li 

Now  subtracting  (2)  from  this  last  equation  and  reducing  we  obtain 

W  W 

dM=Wdz--^-  adz-  —  2zdz  ..........................  (3) 

Li  Li 

which  is  undoubtedly  the  expression  for  the  increment  of  the  bending 
moment  under  wheel  C  due  to  any  slight  movement  of  the  loads.  It  is 
readily  seen  that  if  wheel  C  is  to  the  right  of  the  point  of  maximum 
moment  any  slight  movement  of  the  loads  to  the  left  will  increase  the 
moment  M  by  the  amount  dM  —  in  other  words,  dM  will  be  positive  —  and 
that  all  such  movements  of  the  loads  to  the  left  will  increase  M  the  amount 
dM  until  wheel  C  arrives  at  the  point  of  maximum  moment  and  from  there 
on  any  slight  movement  of  the  loads  to  the  left  will  decrease  M  by  the 
amount  dM;  in  other  words,  dM  is  negative  in  that  case.  Evidently  the 
increment  dM  is  zero  just  as  wheel  C  arrives  at  the  point  of  maximum 
moment  as  it  is  positive  when  the  wheel  is  just  to  the  right  of  the  point 
of  maximum  moment  and  negative  when  to  the  left  and  hence  changes 
signs  at  the  point.  So  we  have  from  (3),  when  M  is  a  maximum, 

dM  =  Wdz-~  adz-~2zdz  =  0. 
Li  Li 

Reducing,  we  obtain 


which  is  the  value  of  z  when  the  maximum  bending  moment  occurs  under 
wheel  C.  This  last  equation  shows  that  the  center  of  the  beam  bisects 
the  distance  a.  From  this  we  see,  that  the  point  of  maximum  moment  and 
the  center  of  gravity  of  all  the  wheels  are  equidistant  from  the  center  of 
the  beam  and  on  opposite  sides  of  the  center.  The  maximum  moment  will 
occur  under  one  of  the  two  wheels  adjacent  to  the  center  of  gravity, 
practically  always  under  the  wheel  nearer  the  center  of  gravity.  So,  as  a 
rule,  all  we  have  to  do  to  locate  a  system  of  wheels  on  a  beam  for 
maximum  moment  is  to  find  their  center  of  gravity  and  then  place  the 
center  of  the  beam  half  way  between  this  center  of  gravity  and  the  wheel 
nearest  to  it.  For  example,  take  the  case  shown  in  Fig.  112.  By  taking 
moments  about  either  of  the  end  wheels,  either  wheel  D  or  E,  we  can 
determine  the  center  of  gravity  of  all  the  wheels  which  comes  nearest  to 
wheel  C  and  hence  the  center  of  the  beam  will  come  half  way  between 
the  center  of  gravity  and  this  wheel.  In  any  case  where  the  center  of 
gravity  comes  near  half  way  between  two  wheels,  the  moment  under  each 
of  the  wheels  should  be  determined  in  order  to  ascertain  which  is  the 
maximum. 


130 


STRUCTURAL  ENGINEERING 


Usually  the  only  bending  moment  used  in  designing  beams  is  the 
absolute  maximum  referred  to  above,  yet  there  are  a  few  cases  where  the 
maximum  at  other  points,  other  than  the  point  where  the  absolute  maxi- 
mum occurs,  is  needed,  and  which  will  occur  when  the  wheels  are  in  one 
certain  position  in  reference  to  the  point  considered. 

Let  AB  (Fig.  113)  represent  a  simple 
beam  supporting  a  system  of  wheel  loads 
as  shown,  and  let  D  be  a  point  distance  e 
from  A  where  the  maximum  bending  mo- 
ment, due  to  these  wheels  as  they  move 
over  the  beam,  is  desired.  Let  W  repre- 
ss sent  the  weight  of  all  of  the  wheels  on  the 
beam  and  x  the  distance  of  their  center  of 
gravity  from  B.  Further,  let  P  represent 
the  weight  of  the  wheels  to  the  left  of 
point  D  and  2  the  distance  of  their  center 
of  gravity  from  D. 

According  to  Art.  56,  the  maximum  moment  at  D  will  occur  when 
a  wheel  is  at  that  point.  So  let  a  wheel  be  at  the  point  as  shown.  Now 
taking  moments  about  D  and  considering  the  forces  to  the  left,  we  have 

W 


e 

L-e 

0  0 

z 

\ 

oc 

( 

OOO 

If 

R 

'P 

Fig 

u 
\ 
r,  i 

/ 

V 

13 

JRI 

=  —  ex-Pz=(Re-Pz) 


for  the  bending  moment  at  that  point.  Now,  if  the  wheels  are  in  the 
position  for  maximum  moment  at  D,  the  increment  of  the  moment  that 
would  be  caused  by  the  wheels  moving  to  the  right  or  left  an  infinitesimal 
distance  would  be  equal  to  zero.  Then  assuming  we  have  such  a  move- 
ment, M,  x,  and  s  will  have  corresponding  increments  which  we  will 
designate,  respectively,  as  dM,  dx,  and  dz.  Now  adding  these  in  equation 
(1),  we  have 

W         W 


and  subtracting  equation  (1),  we  have 


But,  as  is  readily  seen,  dz  —  dx,  so  we  have 


for  the  increment  of  the  bending  moment,  which  will  be  equal  to  zero 
when  the  loads  are  in  the  position  for  maximum  moment  at  D.  Then  we 
have 


Reducing,  we  have 


W     P 
~L=7 


(2), 


MAXIMUM  REACTIONS,  ETC. 


131 


In  the  last  equation  we  have  the  total  load  on  the  beam  divided  by 
the  length  of  the  beam  equal  to  the  load  on  the  left  of  the  point  D 
divided  by  the  distance  from  the  point  to  the  left  support.  That  is,  we 
have  the  average  unit  load  on  the  beam  equal  to  the  average  unit  load  to 
the  left  of  the  point.  Then,  to  obtain  the  maximum  moment  at  any  point 
in  a  simple  beam,  due  to  wheel  loads,  the  loads  must  be  so  placed  that 
there  will  be  a  load  at  the  point  considered  and  at  the  same  time  the 
average  unit  load  on  the  entire  beam  will  be  equal  to  the  average  unit 
load  to  the  left  of  the  point. 

It  is  shown,  in  Art.  59,  that  the  first  derivative  of  the  bending 
moment,  in  terms  of  x,  at  any  section  of  a  beam,  is  equal  to  the  shear. 
But  at  the  point  of  maximum  moment  the  first  derivative  of  the  bending 
moment  will  be  equal  to  zero.  So,  it  follows  that  the  shear  at  the  point 
of  maximum  bending  moment  will  be  zero.  Then,  as  the  point  of  maxi- 
mum bending  moment  is  at  the  point  of  zero  shear,  or  where  the  shear 
changes  signs,  which  is  the  same  thing,  the  point  of  maximum  moment 
can  be  obtained  by  adding  up  the  forces  from  one  end  of  the  beam  until 
the  point  where  the  shear  is  zero,  or  changes  signs,  is  reached,  and  this 
point  will  be  the  point  of  maximum  bending  moment.  This  method  of 
determining  the  point  of  maximum  moment  is  quite  convenient  in  the  case 
of  fixed  loads,  especially  where  the  loads  are  mixed,  uniform,  and 
concentrated. 

As  an  example,  let  it  be  required  to  determine  the  point  of  maximum 
bending  moment  on  the  beam  shown  in  Fig.  114,  due  to  the  concentrated 
and  uniform  load  indicated.  After  determining  the  reactions  shown  at 
A  and  B,  by  taking  moments  about  the  supports,  we  can  begin  at  one  end 
of  the  beam  and  ascertain  the  point  of  zero  shear  by  adding  up  the  forces 

3'        ._.*'   .      S'  ,..    Z'  .        «?- 

*  T          x 


B 


Fig.   114 

and  noting  the  sign  of  the  shear  as  we  progress.  Thus,  beginning  at  A, 
we  have  10,766- 1,800 -3,000  =  +5,966  Ibs.  for  the  shear  just  to  the  right 
of  the  3,000-pound  load,  and  10,766-3,000  -  7,000  =  +766  Ibs.  for  the 
shear  just  to  the  right  of  the  4,000-pound  load.  This  last  shear  shows  us 
that  the  point  of  zero  shear  is  just  a  little  to  the  right  of  the  4,000-pound 
load.  Let  x  be  the  distance.  Then  we  have 


from  which  we  obtain 


600  x  =  766, 
766 


600 


=  1.27  ft. 


132 


STRUCTURAL  ENGINEERING 


So  the  point  of  maximum  moment  is  1.27  ft.  to  the  right  of  the  4,000- 
pound  load.  If  the  4,000-pound  load  were  something  over  766  pounds 
heavier,  it  is  evident  that  the  point  of  maximum  moment  would  be  under 
that  load.  For  example,  suppose  that  load  weighed  6,000  pounds  instead 
of  4,000 ;  then,  all  the  other  loads  remaining  the  same,  the  reaction  at  A 
would  be  11,932  Ibs.  instead  of  10,766.  Adding  up  from  A  we  have 
11,932 -3,000 -3,000  =+5,932  Ibs.  for  the  shear  just  to  the  left  of  the 
second  load  from  A,  which  is  now  6,000  Ibs.,  and  11,932-3,000-3,000- 
6,000  =  -68  Ibs.  for  the  shear  just  to  the  right  of  the  load.  As  the  shear 
is  positive  on  one  side  of  the  load  and  negative  on  the  other,  the  shear 
undoubtedly  passes  through  zero  at  the  load,  and  hence  the  point  of 
maximum  moment  is  at  the  load. 

89.  Maximum  Reaction  on  Simple  Trusses. —  The  loads  on  trusses 
are  transmitted  to  the  joints,  either  directly  or  indirectly;  directly  when 
applied  at  the  joints,  and  indirectly  when  applied  between  the  joints, 
that  is,  in  the  panels.  As  an  example,  let  the  diagram  in  Fig.  115 


Fig.  115 

represent  a  truss  of  length  L  supporting  two  loads,  PI  and  P2,  as  shown. 
These  loads,  as  is  readily  seen,  would  be  transmitted  to  the  joints  c,  d, 
and  e.  Let  rl,  r2,  and  r3  represent  the  amount  transmitted  to  each  as 
indicated.  These  would  be  known  as  concentrations  or  panel  loads.  The 
intensity  of  rl  and  r3  can  be  determined  by  taking  moments  about  d  and 
treating  cd  and  de  as  simple  beams.  Thus  we  have 

hPl  kP2 

rl  =  — 7-  and  r6  =  — - — 
cd  de 

By  taking  moments  about  e  we  can  determine  the  concentration  at  d  due 
to  P2,  and  by  taking  moments  about  c  we  can  determine  the  concentration 
at  d  due  to  PI.  Then,  by  adding  these  two  concentrations  together,  we 
obtain  the  concentration  r2.  Thus  we  have 


.9- 


cd-h 


P1  + 


de-k 
de 


\P2. 


In  this  manner  the  concentrations  on  the  joints  of  any  truss,  loaded  in 
any  manner,  can  be  determined.  As  the  concentrations  are  really  com- 
ponents of  the  loads,  either  may  be  used  in  determining  reactions  on 
trusses.  Thus,  for  the  reaction  at  A  (Fig.  115),  due  to  the  two  loads 
PI  and  P2,  we  have 


H  — 


ePl  +  yP2 


or  Li  = 


(Bd)r 


MAXIMUM  REACTIONS,  ETC.  133 

As  a  rule,  the  loads  are  used  instead  of  the  concentrations  (owing  to 
convenience),  in  which  case  any  concentration  at  the  end  point  due  to 
loads  in  the  end  panel  must  always  be  subtracted  from  the  reaction  found. 
For  instance,  if  there  were  loads  in  the  panel  Ab,  we  would  take  moments 
about  b  and  determine  the  concentration  at  A  due  to  the  loads  in  that 
panel,  and  subtract  this  from  the  reaction  found  at  A  by  taking  moments 
of  all  the  loads  about  B.  Other  than  this  one  step,  the  determination  of 
reactions  on  trusses  is  exactly  the  same  as  for  beams. 

Dead  load  (which  is  usually  taken  as  a  uniform  load)  and  uniform 
live  load  are  always  considered  as  concentrated  at  the  joints.  In  such 
cases  we  deal  only  with  joint  loads,  or  panel  loads,  which  is  the  same 
thing.  If  the  panels  be  of  equal  length,  the  panel  loads  will  all  be  equal, 
and  in  such  cases  the  reactions  can  be  determined  practically  by  inspec- 
tion. For  example,  let  the  diagram  in  Fig.  116  represent  a  truss  of  6 


equal  panels  supporting  a  uniform  load  of  w  pounds  per  foot  of  truss. 
Let  L  be  the  length  of  the  truss  and  d  the  length  of  each  panel.  If  the 
load  extends  over  the  full  length  of  the  span  there  would  be  five  panel 
loads,  represented  as  P,  each  equal  to  wd.  In  that  case  we  can  readily 
see  that  each  reaction  would  be  equal  to  2J  P.  This,  of  course,  is  mere 
inspection.  In  case  the  panel  loads  are  unequal  or  some  of  the  joints 
not  loaded,  we  can  obtain  the  reactions  by  simply  taking  moments  about 
the  supports  just  the  same  as  in  the  case  of  beams,  except  as  stated  above, 
but  if  the  panels  are  of  equal  length  it  is  more  convenient  to  deal  with 
panel  lengths  instead  of  feet.  For  example,  to  obtain  the  reaction  at  A 
due  to  a  load  at  /,  we  would  take  moments  about  B  and  the  load  at  / 
would  be  multiplied  by  d  and  divided  by  6d,  and  hence  the  reaction  at 
A  would  be  J  of  the  load.  Similarly  a  load  at  e  would  be  multiplied  by 
2d  and  divided  by  Gd,  and  hence  the  reaction  at  A,  due  to  this  load, 
would  be  f  of  the  load  at  e,  and  likewise  the  reaction  at  A,  due  to  a  load 
at  d,  would  be  £  of  the  load  at  d,  and  £  and  f  for  a  load  at  c  and  b, 
respectively.  From  this  it  is  seen  that  the  reactions  on  trusses  of  equal 
panels  can  be  obtained  directly  by  numbering  the  panels  1,  2,  3,  etc.,  as 
shown  (just  below  the  diagram)  in  Fig.  116,  and  multiplying  the  load  on 
each  j  oint  by  the  number  under  it,  and  then  adding  all  the  results  together 
and  dividing  by  the  number  of  panels  in  the  truss.  If  the  reaction  at  the 
other  support  be  desired,  the  numbers  would  simply  be  reversed  in  order. 
If  the  panel  loads  are  all  equal,  the  work  will  be  shortened,  of  course,  as 
the  numbers  under  all  the  joints  loaded  can  be  added  together  and  multi- 
plied by  one  panel  load,  divided  by  the  number  of  panels.  As  an  example, 
suppose  the  joints  /,  e,  d,  and  c  each  support  a  load  P.  The  reaction  at 
A  due  to  these  four  loads  would  be 

?10. 
6 


134  STRUCTURAL  ENGINEERING 

If  the  joint  b  were  loaded  also,  we  would  have 


If  joints  e,  c,  and  b  alone  were  loaded,  we  would  have 

P  P 

^  =  £-(2  +  4  +  5)=-  11. 

If  joint  c  alone  were  loaded,  we  would  have 

P  P 


and  so  on.  In  general,  the  maximum  reaction  on  a  truss,  the  same  as  a 
beam,  will  occur  when  the  span  is  fully  loaded  and  the  heaviest  loads 
are  near  the  support  considered.  Specific  cases  will  be  taken  up  later. 
90.  Maximum  Shear  on  Simple  Trusses.  —  The  determination  of 
the  shear  on  trusses  is  the  same  as  that  of  beams,  except  the  loads  are  con- 
sidered to  be  applied  to  trusses  only  at  the  panel  points,  or  joints.  Let 
the  diagram  in  Fig.  117  represent  a  simple  truss  supporting  a  load  P  at 
each  bottom  joint,  as  indicated,  and  let  R  and  Rl  represent  the  reactions 
due  to  these  loads.  Then  the  shear  in  the  first  panel  from  the  left  end  is 
Rj  in  the  second  it  is  R-P;  in  the  third  it  is  R-2P;  in  the  fourth 
it  is  R  -  3P;  and  so  on.  This  case  is  that  of  dead  load  or  a  uniform  live 


load  extending  all  the  way  across  the  span.  In  case  of  dead  load,  this 
shear  would  be  a  maximum.  It  is  customary  to  assume  that  a  uniform 
live  load  moving  over  a  truss  increases  by  panel  loads,  in  which  case  the 
maximum  shear  in  any  panel  will  occur  when  the  joints  from  one  end  of 
the  truss  up  to  the  panel  considered  are  loaded.  By  loading  from  one 
end  we  obtain,  as  we  may  say,  the  maximum  positive  shear,  and  loading 
from  the  other  end  we  obtain  the  maximum  negative  shear.  For  example, 
one  maximum  shear  would  occur  in  the  panel  be  of  the  truss  shown  in 
Fig.  118,  when  the  joints  e,  d,  and  c  alone  are  loaded,  and  the  other 
maximum  would  occur  when  the  joints  a  and  b  alone  are  loaded.  Instead 
of  considering  the  load  to  move  on  from  the  two  directions,  we  usually 
consider  it  to  move  on  from  only  one  direction,  and  by  determining  the 
shear  in  each  panel  as  the  load  reaches  it  we  obtain  all  the  shears  we 
need,  as  the  positive  shears  are  determined  in  the  panels  on  one  side  of 
the  center  of  the  truss  while  the  negative  shears  are  determined  in  the 
corresponding  panels  on  the  other  side  of  the  center. 

As  the  maximum  shear  in  any  panel,  due  to  a  uniform  live  load, 
occurs  when  the  load  just  reaches  that  panel,  it  is  readily  seen  that  this 
shear,  in  all  cases,  will  be  equal  to  the  reaction  at  the  unloaded  end  of  the 
truss.  Thus,  the  maximum  shear  in  panel  ed  (Fig.  118)  due  to  a  uniform 
live  load  moving  on  from  the  right,  will  be  equal  to  the  reaction  at  A  due 


MAXIMUM  REACTIONS,  ETC. 


135 


to  the  load  at  e;  and,  similarly,  the  maximum  shear  in  the  panel  dc  will 
be  equal  to  the  reaction  at  A  due  to  the  loads  at  e  and  d;  and  the  maxi- 
mum shear  in  panel  cb  will  be  equal  to  the  reaction  at  A  due  to  the  loads 
at  e,  d,  and  c;  and  so  on  for  the  other  panels.  Now  if  the  panels  be  of 
equal  length,  these  reactions  can  be  determined  by  numbering  the  joints 
1,  2,  3,  etc.,  and  adding  these  together  for  the  loaded  joints  and  multiply- 
ing the  sum  by  one  panel  load  divided  by  the  total  number  of  panels  in 
the  truss,  as  explained  in  the  last  article.  As  an  example,  let  the  diagram 
in  Fig.  118  represent  a  truss  of  six  equal  panels,  each  of  length  d  (the 


Fig;   118 


length  of  the  span  then  will  be  L  —  Qd},  and  let  it  be  required  to  deter- 
mine the  maximum  shear  in  each  panel  due  to  a  uniform  live  load  of  w 
pounds  per  foot  of  truss  moving  over  it  from  right  to  left.  Let  W  (=wd) 
represent  the  panel  load.  Then  numbering  the  joints  1,  2,  3,  etc.,  from 
right  to  left,  as  shown,  we  have 


for  the  maximum  shear  in  the  panel  ed,  when  joint  e  alone  is  loaded, 


for  panel  dc  when  joints  e  and  d  are  loaded, 


for  panel  cb  when  joints  e,  d,  and  c  are  loaded, 

W 


6 


+  2  +  3  +  4) 


for  panel  ba  when  joints  e,  d,  c,  and  b  are  loaded,  and 


for  panel  aA  when  joints  e,  d,  c,  b,  and  a  are  loaded. 

From  this  it  is  seen  that  the  maximum  shear  on  any  truss  of  equal 
panels,  due  to  a  uniform  live  load  moving  over  the  truss,  can  be  expressed 
by  the  general  formula, 


+O-1)], 


where  n  is  the  number  of  panels  in  the  truss  considered. 

It   will  be   found  quite  convenient  to  write  the  summation  of  the 


136 


STRUCTURAL  ENGINEERING 


numbers  in  the  parentheses  under  the  corresponding  panel  points  as 
shown  in  Fig.  118  where  the  incircled  numbers  are  the  respective  summa- 
tions. These  summations,  as  is  readily  seen,  can  be  made  in  any  case  by 
simply  referring  directly  to  the  diagram  of  the  truss  considered.  Then 
the  shear  in  the  panels  can  be  read  off  of  a  slide  rule  directly  as, 

W 


5(3),      E 

^  TO 


• 

and  so  on. 

In  the  case  of  uniform  live  load,  just  considered,  we  usually  assume, 
as  stated  above,  that  the  load  is  applied  to  the  joints  of  a  truss  in  maxi- 
mum panel  loads.  This  is  not  possible,  however,  as  such  loads  are  con- 
tinuous and  must  necessarily  extend  beyond  the  last  loaded  joint  in  order 
to  fully  load  it.  For  example,  to  fully  load  the  joints  e  and  d  (Fig.  118) 
the  uniform  live  load  would  really  have  to  extend  from  B  to  the  joint  c, 
as  is  readily  seen.  However,  as  will  be  shown  later,  the  maximum  shear 
would  occur  in  panel  dc  when  the  load  extended  a  short  distance  z  beyond 
joint  d.  In  that  case  there  would  be  less  than  a  full  panel  load  at  d  and 
a  small  concentration  at  c  due  to  the  load  in  the  panel  dc,  and  the  shear 
in  the  panel  dc  would  really  be  equal  to  the  reaction  at  A  minus  the 
concentration  at  c.  The  error  resulting  from  the  assumption  stated  above 
is  small  in  the  case  of  a  uniform  live  load,  as  will  be  seen  later,  and  hence 
the  assumption  in  that  case  is  permissible,  but  in  the  case  of  wheel  loads 
no  such  assumption  can  be  made  as  any  slight  shifting  of  the  loads  may 
materially  change  the  shear  in  a  panel  and  consequently  it  is  necessary 
for  us  to  determine  the  exact  position  of  such  loads  for  maximum  shear. 
As  a  general  case,  let  the  diagram  in  Fig.  119  represent  a  truss 


) O O\ A o o  r  /h  O O O O 


Fig.   119 


supporting  a  system  of  wheel  loads.  Now  let  it  be  required  to  determine 
the  position  of  these  wheels  for  maximum  shear  in  panel  be.  Let  W  be 
the  weight  of  all  the  wheels  and  x  the  distance  from  B  to  their  center  of 
gravity,  and  let  P  be  the  weight  of  the  loads  in  the  panel  be,  and  z  the 
distance  from  c  to  their  center  of  gravity.  Now  if  R  be  the  reaction  at  A 
due  to  W,  and  r  the  concentration  at  b  due  to  P,  the  shear  in  the  panel 
be  can  be  expressed  as 


MAXIMUM  REACTIONS,  ETC.  137 

=  R-r   ...........................................  (1). 

W  r 


But 

L  a 

Then  substituting  these  values,  we  have 


With  P  and  W  constant,  any  shifting  of  the  wheels  to  the  right  or  left 
will  affect  R  and  r,  not  the  same,  but  similarly.  The  weight  of  P  could 
be  such  that  any  movement  to  the  left  would  increase  the  value  of  R 
more  than  r.  Then,  evidently,  any  such  movements  to  the  left  would 
continue  to  increase  the  shear  S  until  another  load  rolled  past  c  into  the 
panel  be,  when  the  rate  of  increase  of  r  would  be  greater  than  just  before 
the  load  passed  joint  c;  but  R  would  continue  to  increase  all  the  while, 
and,  consequently,  the  shear  S  would  continue  to  increase,  unless  the  load 
P  were  increased  so  much  by  the  additional  load  that  r  would  be  increased 
at  a  greater  rate  than  R  for  each  movement  to  the  left.  So  it  is  seen  that 
the  shear  S  will  be  a  maximum  when  the  total  load  in  the  panel  be  is  such 
that  the  increment  of  R  is  equal  to  the  increment  of  r.  That  is,  any 
shifting  of  the  loads  with  W  and  P  constant  (for  that  instant)  would  not 
affect  the  shear  S  in  panel  be;  or,  in  other  words,  the  increment  of  the 
shear  would  be  zero.  Now,  suppose  the  loads  move  an  infinitesimal 
distance  to  the  left;  then  equation  (2)  would  become 

Wdx     Pz     Pdx 


and  subtracting  (2)  from  this,  we  have 


which  is  the  increment  of  the  shear  S  in  panel  be  due  to  the  movement. 
But  this  increment  will  be  zero  when  S  is  a  maximum,  and  then  we 
would  have 


It  is  readily  seen  that  this  occurs  when  W/L  =  P/d,  that  is,  the  shear  in 
panel  be  is  a  maximum  when  the  average  unit  load  on  the  truss  is  equal  to 
the  average  unit  load  in  the  panel.  This  will  hold  in  the  case  of  any  truss 
of  the  type  shown  as  the  case  being  considered  is  general.  In  the  case  of 
a  truss  of  n  equal  panels,  each  of  length  d,  we  have  L  =  nd.  Substituting 
this  value  of  L  in  the  last  equation,  and  reducing,  we  have 


That  is,  the  shear  in  any  panel  of  a  truss  of  n  equal  panels  will  be  a 
maximum  when  the  load  in  the  panel  is  equal  to  the  total  load  on  the 
truss  divided  by  the  number  of  panels  in  the  truss.  This  will  hold  for 
any  class  of  live  load  either  concentrated  or  uniform. 

It   is   readily   seen   that   the   increment   of  the   shear   in   panel   be, 
expressed  by  equation  (3),  will  pass  through  zero  only  as  a  load  passes 


138 


STRUCTURAL  ENGINEERING 


joint  c,  so  there  will  be  a  wheel  at  joint  c  when  the  maximum  shear  in 
panel  be  occurs.  Thus  it  is  in  all  cases — a  wheel  will  be  at  the  joint  on 
the  loaded  side  of  the  panel  considered  when  the  maximum  shear  in  the 
panel  occurs. 

91.  Maximum  Bending  Moments  on  Simple  Trusses. — In  the 
case  of  bending  moments  on  trusses,  the  moments  are  usually  taken  about 
the  joints  only,  otherwise  the  case  is  the  same  as  that  of  beams.  As  a  gen- 
eral case  of  dead  load  or  uniform  live  load,  let  the  diagram  in  Fig.  120 
represent  a  truss  of  six  panels  supporting  a  load  at  each  bottom  joint  as 


q  . 

i 
h           \ 

\ 

\ 

\k 


PI 


c\         a 

p^       PS 


Fig.   120 

indicated.  Let  R  and  Rl  represent  the  reaction  at  A  and  B,  respectively, 
due  to  these  loads.  Then  the  bending  moment  at  joint  b  or  g  is  M'  =  Rx; 
at  joint  c  or  h  it  is  M"=R(x  +  y)-Ply ;  at  joint  k  or  d  it  is  M"  f- 
R(x  +  y  +  z)-Pl(y  +  z)-P2(z);  and  so  on. 

In  the  case  of  uniform  live  load,  the  maximum  moment  at  any  joint 
will  occur  when  the  load  extends  the  full  length  of  the  truss,  and  hence 
the  moment  at  all  joints,  the  same  as  in  the  case  of  dead  load,  is  a 
maximum.  In  the  case  of  wheel  loads  it  is  necessary  to  determine  the 
exact  position  of  the  wheels  for  the  maximum  bending  moment  at  each 
joint.  Let  the  diagram  in  Fig.  121  represent  a  truss  of  six  panels 
supporting  a  system  of  wheel  loads,  as  indicated.  Now  let  it  be  required 
to  determine  the  position  of  these  wheels  for  maximum  bending  moment 
at  joint  d. 

Let  W  be  the  weight  of  all  the  wheels  on  the  truss,  and  x  the 
distance  from  B  to  their  center  of  gravity,  and  let  P  be  the  weight  of  the 


Fig.   121 

wheels  to  the  left  of  joint  d  and  z  the  distance  from  d  to  their  center  of 
gravity,  and  also  let  k  be  the  distance  from  A  to  d  and  R  the  reaction 
at  A  due  to  all  the  wheels  when  in  any  position.  Then  the  bending 
moment  at  d  can  be  expressed  as 

M  =  Rk-Px  ..,,  (1). 


MAXIMUM  REACTIONS,  ETC.  139 

But  |  B=*i  - 

LJ 

so,  substituting  this  value  of  R,  we  have 

W  r 

M  =  k~-Pz   .....................................  (2). 

Now  as  the  loads  move  to  the  left,,  R  will  be  increased,  and  likewise 
the  moment  M,  until  P  becomes  so  great  that  the  rate  of  increase  of  Pz 
is  greater  than  the  rate  of  increase  of  (Wx  /  L)k;  then,  evidently,  the 
moment  M  at  d  will  be  a  maximum  when  its  increment  is  zero,  which,  as  is 
readily  seen,  will  occur  when  the  increment  of  (Wx  /  L)k  is  equal  to  the 
increment  of  Pz.  Now  suppose  the  loads  move  an  infinitesimal  distance  to 
the  left,  then  equation  (2)  becomes 


Then  subtracting  (2)  we  have 


which  is  the  increment  of  the  bending  moment  at  d  due  to  the  movement. 
But  this  increment  will  be  equal  to  zero  when  M  is  a  maximum,  and  hence 
we  would  have 


(4) 


It  is  readily  seen  that  this  occurs  when 


that  is,  the  bending  moment  at  joint  d  is  a  maximum  when  the  average 
unit  load  on  the  truss  is  equal  to  the  average  unit  load  on  the  left  of  the 
joint  d.  Now,  as  the  above  is  a  general  case,  we  have:  The  maximum 
bending  at  any  joint  of  a  truss  of  the  type  shown  will  occur  when  the 
average  unit  load  on  the  truss  is  equal  to  the  average  unit  load  to  the  left 
of  the  joint  considered.  It  will  be  seen  that  this  is  the  same  as  found  for 
beams  in  Art.  88. 

92.  Stresses  in  Trusses.  —  As  the  loads  supported  by  trusses  are 
applied  at  the  joints,  the  stress  produced  in  each  member  is  simple 
stress,  and  consequently  the  unit  stress  in  any  member  is  equal  to  the 
total  stress  divided  by  the  area  of  its  cross  section.  The  stresses  in  the 
individual  members  can  readily  be  obtained  by  resorting  to  the  resolution 
of  forces  and  to  the  equation  of  moments. 

As  an  example,  let  the  diagram  at  (a)  in  Fig.  122  represent  a  truss 
of  six  equal  panels,  supporting  a  load  at  each  lower  joint  as  indicated. 
Let  d  be  the  length  of  each  panel,  h  the  height  of  the  truss  center  to 
center  of  chords,  L  the  total  length  of  span  center  to  center  of  end 
bearings,  and  let  R  and  Rl  represent  the  reactions  as  indicated.  To 


140 


STRUCTURAL  ENGINEERING 


obtain  the  stress  in  the  end  post  LOU1  and  in  the  bottom  chord  LOL1, 
let  S  and  SI  be  the  stress  in  each,  respectively,  and  suppose  the  two 
members  cut  off  along  the  section  mm,  and  imagine  the  part  of  the  truss 
to  the  left  of  this  section  moved  bodily  to  (6)  without  any  stresses  or 
forces  being  affected  in  the  least.  Then  we  simply  have  an  independent 


m"' 


Fig.   122 

structure  at  (&)  held  in  equilibrium  by  the  three  external  forces  R,  S, 
and  SI  —  the  stresses  S  and  $1  being  unknown.  Now,  as  these  three 
forces  are  in  equilibrium,  the  algebraic  sum  of  their  components  along 
any  direction  is  equal  to  zero.  So  resolving  the  forces  vertically,  we  have 

0   .......................................  (1), 


from  which  we  obtain 


(2)  ; 
and  resolving  the  forces  horizontally,  we  have 

SI  -Ss'm6  =  0    ......................................  (3), 

from  which  we  obtain 

SI  =  SsinO    .........................................  (4). 


MAXIMUM  REACTIONS,  ETC.  141 

But,  according  to  (2),  S  =  R/cos6.     Then,  substituting  in  (4),  we  have 

S1=R  -^  =  Rtan6    ................................  (5). 

cos<9 

Equation  (2)  shows  that  the  stress  S  in  LOUl  is  equal  to  the  shear 
(=#)  in  the  end  panel  multiplied  by  the  secant  of  the  angle  6  which  the 
member  makes  with  the  vertical;  and  equation  (5)  shows  that  the  stress 
SI  in  LOZ/1  is  equal  to  the  shear  in  the  end  panel  multiplied  by  the 
tangent  of  angle  6.  It  should  be  observed  that  R  has  no  horizontal  com- 
ponent while  $1  has  no  vertical  —  the  cosine  of  90Q  being  zero. 

Again  referring  to  the  figure  at  (b),  by  taking  moments  about  any 
point  O  on  the  line  of  action  of  S,  we  have 

Rx 
from  which  we  obtain 


But  it  is  obvious  that  it  is  more  convenient  to  take  moments  about  the 
panel  point  £71,  for  in  that  case  the  lever  arms  are  directly  known,  and 
we  have 

Rd-hSl  =  0, 
from  which  we  obtain 


(6). 


The  stress  S  could  be  computed  by  taking  moments  about  a  point  in  the 
line  of  action  of  the  stress  £1,  but  it  is  obvious  that  S  can  be  computed 
more  readily  from  equation  (2). 

It  is  also  obvious  that  the  stress  in  the  hanger  U1LI  is  equal  to  the 
load  P  which  it  supports  directly  at  LI. 

To  obtain  the  stress  in  the  chords  £71  £72,  L1L2,  and  diagonal  £71L2, 
let  S2,  S3,  and  $4  be  the  stress  in  each,  respectively,  and  suppose  the 
three  members  cut  off  along  the  section  m'm'  ,  and  imagine  the  part  of 
the  truss  to  the  left  of  this  section  moved  bodily  to  (c)  without  any 
stresses  or  forces  being  affected  in  the  least.  Then  we  have  an  inde- 
pendent structure  at  (c)  held  in  equilibrium  by  the  external  forces  R,  P, 
S2,  S3,  and  £4  —  the  stresses  S2,  S3,  and  S4  being  unknown. 

As  S2  and  £4  intersect  at  £71,  S3  can  be  determined  by  taking 
moments  about  that  point.  So  taking  £71  as  the  center  of  moments 
we  have 

Rd-hS3  =  0, 
from  which  we  obtain 


(7), 


which  is  the  same  as  we  obtained  for  the  stress  in  LOL1  in  equation  (6). 
As  £4  and  S3  intersect  at  L2,  S2  can  be  determined  by  taking  moments 
about  L2.  Therefore,  taking  L2  as  the  center  of  moments,  we  have 

2dR-dP-hS2  =  Q 


142  STRUCTURAL  ENGINEERING 

from  which  we  obtain 

S2  =  2^R~^P    (8). 

£4,  the  stress  in  the  diagonal  £71L2,  can  be  determined  most  readily 
from  the  algebraic  sum  of  the  vertical  components  of  the  five  forces,  for 
which  we  have 


from  which  we  obtain 

(9)- 


This  shows  that  the  stress  $4  in  the  diagonal  U1L2  is  equal  to  the  shear 
in  the  panel  L1L2  multiplied  by  the  secant  of  angle  6. 

To  obtain  the  stress  in  the  vertical  post  U2L2,  let  £6  be  the  stress, 
and  suppose  the  three  members  £71172,  U2L2,  and  L2L3  be  cut  off  along 
the  section  m"m"  ,  and  imagine  the  part  of  the  truss  to  the  left  of  this 
section  moved  bodily  to  (e)  without  any  forces  or  stresses  being  affected 
in  the  least.  Then  we  have  an  independent  structure  at  (e)  held  in 
equilibrium  by  six  external  forces—  #,  £7,  £6,  S5,  P,  and  PI.  The 
stresses  S7,  S6,  and  S5  may  all  be  unknown.  As  *S7  and  S5  have  no 
vertical  components  (being  horizontal  members),  S6  can  be  determined 
directly  from  the  algebraic  sum  of  the  vertical  components  of  all  the 
forces,  for  which  we  have 

R-2P-S6  =  0, 
from  which  we  obtain 

S6  =  R-2P   .......................................  (10). 

This  shows  that  £6  is  equal  to  the  shear  in  the  panel  L2L3. 

To  obtain  the  stress  in  the  chords  £72  £73,  L2L3,  and  the  diagonal 
£72L3,  let  $10,  £8,  and  £9  be  the  stress  in  each,  respectively,  and  sup- 
pose the  three  members  cut  off  along  the  section  m"  'mff  ',  and  imagine 
the  part  of  the  truss  to  the  left  of  this  section  moved  bodily  to  (/)  with- 
out any  forces  or  stresses  being  affected  in  the  least.  Then  we  have  an 
independent  structure  at  (/)  held  in  equilibrium  by  six  forces,  R,  *S10, 
S9,  SS,  P,  and  PI,  the  stresses  S10,  59,  and  S8  being  unknown.  Taking 
moments  about  £72,  we  have 

2dR-dP-hS8  =  Q, 
from  which  we  obtain 


Then  taking  moments  about  L3,  we  have 

3dR  -  2dP  -  dPl  -  hSIQ  =  0, 
from  which  we  obtain 


(11). 


(12). 


MAXIMUM  REACTIONS,  ETC.  143 


Resolving  the  forces  vertically,  we  have 

R-P-Pl-S9cos6  = 
from  which  we  obtain 


COSC7 


=sec0(J2-P-Pl)    ....................  (13). 


The  above  loads  do  not  produce  any  stress  in  the  post  U3L3.  This 
member  could  be  omitted  as  far  as  loads  applied  at  the  lower  panel  points 
are  concerned,  but  it  will  support  directly  any  load  applied  at  [73. 

Stress  in  the  members  to  the  right  of  C73L3  can  be  determined  in 
the  same  manner  as  shown  above  for  the  members  to  the  left. 

The  above  shows  how  readily  the  stresses  in  a  truss  can  be  deter- 
mined by  considering  portions  of  the  truss  as  independent  structures,  the 
portion  being  selected  each  time  to  suit  the  case  considered;  the  portion 
may  be  a  single  joint  or  several  panels  of  a  structure.  This  manner  of 
treatment  is  known  as  the  "Method  of  Section."  It  will  apply  in  the  case 
of  any  truss  and  may  be  carried  to  any  desired  extent,  but  after  one 
becomes  familiar  with  a  certain  type  of  truss,  the  analysis  is  often  made 
without  any  thought  of  the  method  of  section,  as  will  be  seen  later. 
However,  the  method  is  really  implied. 

Problem  1.  Determine  the  stresses  in  the  members  of  the  truss 
shown  at  (a),  Fig.  120,  due  to  a  load  of  20,000  Ibs.,  on  each  lower  panel 
point,  assuming  the  length  of  each  panel  to  be  25'-0",  and  the  height  of 
the  truss  30'-0",  center  to  center  of  chords. 


CHAPTER  VIII 

GRAPHIC  STATICS 

93.  Definition  and  Limitation   of  Graphic   Statics. —  Graphic 

Statics  is  that  part  of  Mechanics  which  treats  of  the  graphical  determina- 
tion of  static  forces.  It  results  from  the  fact  that  forces  acting  upon  a 
body  in  equilibrium  will  form  a  closed  polygon  (see  Art.  42)  wherein  all 
the  forces  act  in  the  same  direction  around  the  polygon.  This  knowledge 
applied  to  geometrical  construction  constitutes  the  method. 

A  force  is  fully  known  when  its  point  of  application,  its  intensity, 
the  direction  and  location  of  its  line  of  action,  and  its  direction  of  action 
(along  the  line  of  action)  are  known. 

The  determination  of  the  point  of  application  does  not  come  within 
the  scope  of  Graphic  Statics,  for  it  is  one  of  the  many  things  that  can  be 
ascertained  only  from  knowledge  of  actual  conditions,  and  consequently 
is  a  presupposed  knowledge  in  graphic  problems.  The  intensity,  direc- 
tion, and  location  of  the  line  of  action  and  the  direction  of  action  can  be 
graphically  determined.  However,  the  method  is  practically  limited  to 
the  two  following  cases: 

Case  I.  The  intensity  and  direction  of  action  of  two  forces  can  be 
graphically  determined  provided  all  of  the  other  forces  acting  upon  the 
same  body  are  fully  known  and  the  lines  of  action  of  the  two  are  known. 
This  will  hold  in  the  case  of  one  unknown  force  as  well  as  for  two. 

Case  II.  The  intensity,  direction  of  action,  and  the  direction  and 
the  location  of  the  line  of  action  of  one  force  can  be  graphically  deter- 
mined provided  all  of  the  other  forces  acting  upon  the  same  body  are 
fully  known. 

Concisely  expressed,  the  intensity  and  direction  of  action  of  two 
forces  can  be  found  when  the  lines  of  action  of  all  the  forces  are  known 
and  the  intensity  and  direction  of  action  of  all  except  those  two  are 
known ;  and  the  intensity,  direction  of  action,  and  the  direction  and 
location  of  the  line  of  action  of  one  force  can  be  found  when  all  the  other 
forces  are  known.  As  a  rule,  the  intensity  and  direction  of  action  are  all 
that  are  desired. 

94.  Preliminary  Application. — Let  AB,  at  (a),  Fig.  123,  represent 
a  body  in  equilibrium  under  the  action  of  the  five  forces  PI — P5.     Sup- 
pose all  of  the  forces  are  known  except  P4  and  P5,  which  are  unknown 
in  intensity  and  direction  of  action  only,  which  means  their  lines  of  action 
are  known,  so  that  the  problem  comes  under  Case  I. 

To  determine  their  intensity  and  direction  of  action,  select  some 
convenient  scale  and  lay  off  AB  at  (6),  equal  and  parallel  to  PI;  and 
from  B  lay  off  BC  equal  and  parallel  to  P2 ;  and  from  C  lay  off  CD  equal 
and  parallel  to  P3.  Then,  to  complete  the  polygon,  P4  and  P5  must 
close  on  D  and  A.  So  from  D  draw  Dy  parallel  to  P4  and  from  A  draw 
A z  parallel  to  P5,  intersecting  Dy  at  E.  Then  the  length  of  the  lines 
DE  and  AE  represents  the  intensity  of  P4  and  P5,  respectively.  Now,  as 

144 


GRAPHIC  STATICS 


all  of  the  forces  will  act  in  the  same  direction  around  the  polygon.  P4 
will  act  in  the  direction  from  D  to  E  and  P5  from  E  to  A,  as  the  arrow 
points  indicate.  So  we  have  their  direction  of  action,  which  is  toward 
the  body  in  each  case.  Then  the  forces  P4  and  P5  are  fully  known.  The 
two  forces  P4  and  P5  could  close  on  A  and  D,  as  AO  and  DO,  just  as 
well  as  DE  and  AE.  Either  is  correct. 

It  is  usually  convenient  to  go 
around  a  body  laying  off  the  forces 
in  consecutive  order  as  was  done  at 
(6),  but  in  some  cases  it  is  not  pos- 
sible to  do  so,  and  in  fact  it  is  not 
at  all  necessary.  For  example,  in 
the  above  case  we  could  lay  off  the 
forces  as  they  are  shown  at  (c)  by 
laying  off  PI  first,  then  P3,  then  P2, 
and  close  on  K  and  H  with  P4  and 
P5 ;  or  we  could  lay  off  either  P2  or  y 
P3  first,  just  so  the  forces  are  joined  / 
so  that  they  act  in  the  same  direc-  * 
tion  around  the  polygon,  which  is  the 
important  thing  to  watch.  However, 
any  mistake  in  that  respect  is  readily 
detected.  For  instance,  suppose  we 
were  to  lay  off  PI  as  AB  (at  d)  and 
then  lay  off  P3  from  A;  we  could 
readily  see  that  the  forces  would  not 
be  acting  in  the  same  direction 
around  the  polygon  being  constructed,  and  consequently  the  polygon 
would  not  be  correct. 

Now  suppose  P2  and  P4  at  (a)  are  the  two  forces  unknown  in 
intensity  and  direction  of  action  instead  of  P4  and  P5  as  considered 
above.  In  that  case  there  would  be  a  known  force  in  between  the  two 
unknown  forces,  and  consequently  it  would  be  impossible  to  lay  off  the 
forces  in  consecutive  order  as  we  pass  around  the  body.  However,  the 
case  presents  no  trouble  at  all,  as  we  can  simply  lay  off  the  known  forces 
in  any  order,  so  long  as  they  are  placed  so  that  they  all  act  in  the  same 
direction  around  the  polygon.  For  example,  to  determine  the  intensity 
and  direction  of  action  of  P2  and  P4,  we  could  lay  off  P5  as  AB  at  (e), 
then  from  B  lay  off  PI  as  EC,  and  from  C  lay  off  P3  as  CD,  then  closing 
on  A  and  D  with  P4  and  P2  by  drawing  Dx  parallel  to  P4  and  Ay 
parallel  to  P2,  intersecting  Dx  at  N.  Then  DN  and  NA  represent  the 
intensity  of  P4  and  P2,  respectively,  and  following  around  the  polygon 
from  A  to  B,  C,  etc.,  it  is  readily  seen  that  P4  acts  from  D  to  N,  and 
P2  from  N  to  A. 

It  is  evident  that  if  only  one  of  the  five  forces  at  (a)  were  unknown 
in  intensity  and  direction  of  action,  our  problem  would  be  quite  simple. 
For  example,  suppose  P5  were  the  only  unknown  force.  We  would  simply 
construct  a  polygon  as  ABCDE,  as  shown  at  (6),  and  join  E  to  A  and 
P5  would  thus  be  determined  in  intensity  and  direction  of  action. 

As  an  example,  under  Case  II,  suppose  all  of  the  forces  shown  at 
(a),  Fig.  123,  are  completely  known  except  P5,  which  we  will  assume  is 


Fig.   123 


146 


STRUCTURAL  ENGINEERING 


completely  unknown.  By  constructing  the  force  polygon  ABCDEA  at 
{f)  we  have  P5  given  in  intensity,  direction,  and  direction  of  action  by 
rhe  line  EA,  but  the  location  of  its  line  of  action  is  unknown.  To  locate 
its  line  of  action  select  any  point  0,  at  (/),  as  a  pole,  and  draw  the  ray 
diagram  as  shown,  and  construct  at  (a)  the  equilibrium  polygon  fabcdf 
(as  explained  in  Art.  45).  Then  prolonging  the  segments  df  and  fa 
until  they  intersect  at  7  and  through  I  drawing  Im  parallel  to  EA,  we 
have  the  line  of  action  of  the  resultant  of  the  four  forces  PI,  P2,  P3, 
and  P4.  Now,  as  all  five  of  the  forces  are  in  equilibrium,  the  resultant 
just  found  must  balance  P5,  which  requires  that  P5  must  act  along  the 
same  line  as  the  resultant.  (See  Art.  42.)  Therefore,  the  line  Im  is 
the  line  of  action  of  P5,  and  thus  P5  is  fully  determined.  P5  may  be 
applied  upon  either  side  of  the  body  as  far  as  equilibrium  is  concerned. 
Referring  to  the  force  polygon  at  (6),  Fig.  123,  it  is  evident  that 
if  P4  and  P5  were  parallel,  a  straight  line  joining  A  and  D  would 
represent  the  direction  and  intensity  of  both,  but  it  would  be  impossible 


W  \ 


Fig.   124 


to  determine  the  intensity  of  either  directly  from  the  polygon,  as  there 
would  be  no  break  in  the  line  indicating  where  the  two  join.  However, 
their  intensities  are  readily  determined  by  the  aid  of  the  equilibrium 
polygon.  For  example,  let  AB,  at  (a),  Fig.  124,  represent  a  body  held 
in  equilibrium  by  the  five  forces  PI.  .  .P5  as  indicated.  Let  PI  and  P5 
be  the  two  unknown  parallel  forces.  By  constructing  the  force  polygon 
ABCDA,  at  (6),  we  have  the  combined  intensities  of  the  two  given  by  the 
line  DA.  Next  select  any  point  as  0,  at  (6),  as  a  pole  and  draw  the  ray 
diagram  ABCDOA.  Then,  beginning  at  any  point  on  the  line  of  action 
of  one  of  the  known  forces,  say  at  b  on  the  line  of  action  of  P2,  construct 
the  equilibrium,  polygon  fbcdf,  as  explained  in  Art.  45.  Then  the  three 
known  forces  P2,  P3,  and  P4  are  replaced  by  components,  all  of  which 
are  balanced  except  /  at  b  and  /'  at  d.  As  the  next  step,  prolong  the 


GRAPHIC  STATICS 


147 


segment  bf  from  b  until  it  intersects  the  line  of  action  of  the  unknown 
force  PI  at  a,  and  likewise  prolong  the  segment  df  from  d  until  it 
intersects  the  line  of  action  of  the  other  unknown  force  Po  at  e.  In  order 
to  extend  the  equilibrium  polygon  farther,  it  is  necessary  to  resolve  PI 
into  two  components  such  that  one  of  them  will  be  equal  and  opposite  to 
the  known  component  /  at  b,  and  P5  into  two  components  such  that  one 
of  them  will  be  equal  and  opposite  to  the  known  component  /'  at  d.  This 
could  be  readily  accomplished  if  the  forces  PI  and  P5  were  known,  but 
as  they  are  unknown  we  have  only  one  component  in  each  case  to  start 
with  ;  that  is,  /  at  a  and  /'  at  e. 

But  as  the  five  forces  are  in  equilibrium,  their  components  will  be  in 
equilibrium.  Therefore,  the  unknown  component  /"  of  PI  must  balance 
the  unknown  component  of  P5,  and  consequently  they  will  act  along  the 
same  line.  So,  by  drawing  the  line  ea,  we  have  the  line  of  action  of  the 
two  unknown  components,  as  this  is  the  only  line  that  both  could  act 
along.  Now  at  e  we  have  one  known  force  /'  and  the  direction  of  the 
two  unknown  forces  Po  and  /"  ',  each  of  which  can  be  graphically  deter- 
mined by  drawing  the  force  triangle  DEO  as  shown  at  (c).  PI  and  /" 
at  a  can  be  graphically  determined  in  the  same  way.  But  it  is  not  neces- 
sary to  construct  the  triangle  at  (c)  as  /'  is  given  in  the  diagram  at  (6) 
by  the  line  OD,  and  by  drawing  OE  parallel  to  the  line  ae  we  would  have 
the  same  forces  given  by  the  equal  triangle  ODE  at  (b),  and  the  forces 
at  a  by  the  triangle  OEA,  as  indicated;  and  thus  PI  and  P5  are  fully 
determined  as  EA  and  ED,  respectively,  in  the  diagram  at  (b). 

Referring  to  Fig.  124,  it  will  be  observed  that  the  equilibrium 
polygon  closes,  which  always  occurs  in  the  case  of  any  system  of  forces 
in  equilibrium.  The  line  ea  is  known  as  the  closing  line. 

95.  Graphical  Determination 
of  Reactions  and  Bending  Mo- 
ments  on  Simple  Beams.  —  Let  AB, 
at  (a),  Fig.  125,  represent  a  simple 
beam  supporting  the  four  vertical 
loads  PI  .  .  .P4  as  indicated,  and  sup- 
pose the  reactions  R  and  Rl,  due  to 
these  loads,  are  unknown  in  intensity. 
First  lay  off  the  load  line  AB  as 
shown  at  (b).  Then  select  any  point 
0  as  a  pole  and  draw  the  ray  dia- 
gram as  shown  and  construct  the 
equilibrium  poly  gonA'abcdB'  at  (c). 
Then  from  O  draw  the  line  OE  par- 
allel to  the  closing  line  A'B',  and, 
according  to  the  last  article,  we  have 
R  given  by  the  line  AE  and  Rl  by 
the  line  BE,  and  thus  the  reactions 
are  determined. 

Conceive  of  the  closing  line  A  'B' 
at  (c)  as  being  a  beam  and  the 

other  part  of  the  equilibrium  polygon  as  a  rope  suspended  from  the  ends 
of  this  beam,  and  imagine  the  loads  supported  upon  the  rope  as  indicated. 
Then  the  equilibrium  polygon  really  becomes  a  structure  upon  which  the 


,  _ 

}    S|  \  S|    \     S|  \ 


\B 


Fig.  125 


148  STRUCTURAL  ENGINEERING 

bending  moment  at  any  vertical  section  will  be  the  same  as  for  the  beam, 
and  consequently  it  may  be  treated  instead  of  the  beam  for  the  purpose 
of  finding  these  moments. 

Then  to  obtain  the  bending  moment  at  the  section  mm  of  the  beam, 
extend  this  section  on  down  to  the  equilibrium  polygon,  cutting  it  off 
along  the  section  ee'  '  .  Then  imagine  the  part  of  the  equilibrium  polygon 
to  the  left  of  this  section  moved  bodily  to  (d)  without  any  loads  or  forces 
being  affected  in  the  least.  Then  taking  moments  about  e,  we  have 

aR-bPl-cP2  =  M. 
Now,  this  moment  is  balanced  by  the  moment  of  f2  about  e,  then  we  have 


and  taking  moments  about  e'  ,  we  have 

M  =  k(f5). 

Both  f2  and  /5  (f2  being  the  stress  in  the  rope  from  b  to  c,  and  /5  the 
stress  in  the  beam  throughout)  are  given  in  the  ray  diagram,  and  d  and  k 
can  be  ascertained  by  scale.  A  better  way  to  obtain  the  moment  is  to 
resolve  f2  and  /5  into  horizontal  and  vertical  components  as  shown,  and 
then  taking  moments  at  either  e  or  e'  '  ,  we  have 


H  is  the  same  for  all  segments,  as  is  seen  from  the  ray  diagram  at  (6), 
and  is  known  as  the  "pole  distance."  Then,  evidently,  the  bending 
moment  at  any  vertical  section  of  the  beam  is  equal  to  this  pole  distance 
(H  )  multiplied  by  the  vertical  ordinate  between  the  closing  line  and  the 
broken  line  of  the  equilibrium  polygon  at  the  same  vertical  section.  For 
example,  the  bending  moment  at  the  section  m'm'  is  equal  to  Hy,  at 
m"m",  Hz,  and  so  on.  The  ordinates  are  measured  in  feet  or  inches  to 
the  same  scale  as  the  beam,  while  H  is  in  pounds,  so  many  thousand 
pounds  per  inch.  In  laying  out  the  ray  diagram,  H  can  always  be  taken 
as  some  convenient  number,  as  100,000  Ibs. 

96.  Graphical  Analysis  of  Trusses  consists  in  graphically  deter- 
mining the  reactions  and  stresses.  The  reactions  are  determined  by 
means  of  the  equilibrium  polygon,  while  the  stresses  are  determined  as 
concurrent  forces  at  the  joints  by  means  of  force  polygons,  one  polygon 
for  each  joint,  the  joints  being  considered  in  the  order  that  the  applica- 
tion of  the  method  requires,  being  usually  the  consecutive  order. 

Example  1.  Let  the  diagram  at  (a),  Fig.  126,  represent  a  truss 
supporting  the  three  loads  PI,  P2,  and  P3,  as  indicated,  and  let  R  and 
Rl  represent  the  reaction  at  A  and  B,  respectively,  due  to  the  three  loads. 

Let  it  be  required  to  determine  the  reactions  on  the  above  truss  and 
the  stresses  in  the  members  due  to  the  three  loads  PI,  P2,  and  P3. 

To  determine  the  reactions,  first  construct  the  ray  diagram  at  (6), 
which  is  accomplished  by  laying  off  the  line  1-2  to  any  convenient  scale, 
equal  and  parallel  to  PI,  and  line  2-3  equal  and  parallel  to  P2,  and  line 
3-4  equal  and  parallel  to  P3,  thus  obtaining  the  load  line  1-2-3-4,  and 
then  taking  any  point  0  as  a  pole  and  drawing  the  rays  1-0,  2-O,  etc. 
Then  construct  the  equilibrium  polygon  CghkD,  which  is  accomplished 


GRAPHIC  STATICS 


149 


by  taking  any  point  on  the  line  of  action  of  one  of  the  forces,  say  point 
g,  on  the  line  of  action  of  PI,  and  resolving  PI  into  two  components  by 
drawing  Cg  parallel  to  the  ray  1-0  and  gh  parallel  to  2-0  and  virtually 
resolving  each  of  the  other  two  loads  into  two  components  by  drawing  hk 
parallel  to  3-0  and  kD  parallel  to  4-0,  and  drawing  the  closing  line  CD 


Fig.   126 


for  completion.  Then  by  drawing  OE  parallel  to  the  closing  line  CD, 
we  have  the  reaction  R  given  by  the  line  1-E  and  Rl  by  the  line  4-£. 
These  lines  can  be  scaled  and  thus  the  reactions  will  be  fully  known  as 
we  know  their  direction  of  action  will  be  upward,  in  order  to  balance  the 
loads. 

After  the  reactions  are  determined  all  of  the  external  forces  are 
known  and  we  can  proceed  to  determine  the  stresses  in  the  truss  members. 

Any  joint  of  any  truss  can  be  considered  as  being  a  small  body 
acted  upon  by  the  truss  members,  meeting  at  the  joint,  and  any  load 
there  applied.  The  details  should  be  such  that  these  will  really  be  con- 
current forces  in  every  case  and  are  so  considered  in  graphic  analysis. 

Then  at  j  oint  A  we  have  three  concurrent  forces :  the  reaction  R,  the 
force  exerted  by  the  member  Ac,  and  the  one  exerted  by  the  member  Ab. 
At  joint  c  we  have  four  concurrent  forces,  the  load  PI  and  one  for  each 
of  the  three  members  meeting  at  that  joint;  at  joint  b  we  have  four  con- 
current forces,  one  for  each  of  the  members  meeting  at  that  joint;  at  joint 
d  we  have  five  concurrent  forces,  the  load  P2  and  one  for  each  of  the 


150  STRUCTURAL  ENGINEERING 

members  meeting  at  that  joint;  and  likewise  at  joints  e,  B,  and  /  we 
have,  respectively,  four,  three,  and  four  concurrent  forces,  as  is  readily 
seen. 

The  intensity  of  the  force  exerted  by  any  member  on  a  joint  will  be 
equal  to  the  stress  in  the  member  exerting  the  force,  and  if  the  force  acts 
toward  the  joint  the  member  exerting  it  will  evidently  be  in  compression 
and  hence  the  stress  in  the  member  will  be  compression,  while  if  the  force 
acts  away  from  the  joint  the  member  exerting  it  will  be  in  tension  and 
hence  the  stress  in  the  member  will  be  tension.  So  if  the  force  exerted 
at  a  joint  by  a  member  be  known,  the  stress  in  the  member  will  be  known. 

Now  it  is  evident  that  the  graphical  determination  of  the  stresses 
in  the  truss  members  is  simply  a  matter  of  determining  the  concurrent 
forces  at  each  joint,  which  we  can  do  by  simply  drawing  a  force  polygon 
for  each  joint.  However,  we  cannot  begin  with  just  any  joint,  as  the 
application  of  the  method  is  limited,  as  stated  in  Art.  86.  It  will  be  seen 
upon  inspection  of  the  diagram  at  (o)  that  the  only  joints  that  can  be 
analyzed  at  the  beginning  are  joints  A  and  B.  Here  we  have,  in  each 
case,  all  forces  known  in  direction  and  only  two  unknown  in  intensity  and 
direction  of  action.  So  we  have  simply  Case  I,  Art.  86.  Then  let  us  take 
joint  A  to  begin  with: 

Laying  off  1-2  at  (c),  to  any  convenient  scale,  equal  and  parallel  to 
R,  and  drawing  2-3  parallel  to  the  member  Ac  and  1-3  parallel  to  the 
member  Ab,  we  have  the  force  diagram  1-2-3-1  representing  the  concur- 
rent forces  at  joint  A,  where  2-3  represents  the  force  exerted  by  the 
member  Ac,  and  1-3  represents  the  force  exerted  by  the  member  Ab. 
By  scaling  2-3  and  1-3  we  obtain  the  intensity  of  these  forces,  respect- 
ively. We  know  that  R  acts  upward  and  hence  will  act  from  1  to  2  in 
the  force  polygon.  Then  as  all  of  the  forces  must  act  in  the  same  direc- 
tion around  the  polygon,  we  have  the  force  exerted  by  Ac  acting  from 
2  to  3  (in  the  polygon)  and  the  one  exerted  by  Ab  acting  from  3  to  1. 
Now  transferring  these  directions  to  joint  A  we  have  the  force  exerted 
by  the  member  Ac  acting  away  from  the  joint  and  the  force  exerted  by 
the  member  Ab  acting  toward  it,  as  indicated.  Then  the  stress  in  Ac  will 
be  tension  and  the  stress  in  Ab  compression,  and  as  the  intensity  of  the 
first  is  given  by  the  line  2-3,  and  the  intensity  of  the  second  by  the  line 
1-3,  in  the  force  polygon,  we  have  the  stresses  in  the  two  members  Ac 
and  Ab  fully  determined. 

Knowing  the  force  exerted  by  the  member  Ac  at  joint  A,  we  know 
the  force  it  exerts  at  joint  c,  as  the  two  will,  undoubtedly,  be  equal  and 
opposite,  and  hence  the  one  at  c  will  act  away  from  the  joint,  as  indicated, 
and  likewise,  knowing  the  force  exerted  at  joint  A  by  the  member  Ab, 
we  know  the  force  it  exerts  at  joint  b,  which  acts  toward  b. 

Joint  b  cannot  be  analyzed  as  yet  for  there  are  still  three  unknown 
forces  acting  at  that  joint,  but  joint  c  can  be,  as  there  are  only  two 
unknown  forces  there.  Then  laying  off  2-3  at  (d)  equal  and  parallel  to 
the  line  2-3  at  (c)  and  2-4  equal  and  parallel  to  PI  and  drawing  4-5 
parallel  to  the  member  cd  and  3-5  parallel  to  cb,  we  have  the  force 
polygon  2-3-5-4-2  representing  the  forces  at  joint  c,  where  the  force 
exerted  by  the  member  cd  is  represented  by  the  line  4-5,  and  the  force 
exerted  by  the  member  cb  is  represented  by  the  line  3-5,  which  is  equiva- 
lent to  saying  that  the  intensity  of  the  stress  in  cd  is  given  by  the  line  4-5, 


GRAPHIC  STATICS 

and  the  intensity  of  the  stress  in  cb  is  given  by  the  line  3-5,,  and  by 
scaling  these  lines  we  have  the  intensity  of  the  stress  in  each  of  the  two 
members.  As  the  force  exerted  at  joint  c  by  the  member  Ac  acts  to  the 
left  from  the  joint,  it  will  act  from  3  to  2  in  the  polygon  at  (d),  and  as 
the  other  forces  must  act  in  the  same  direction  around  the  polygon,  it  is 
readily  seen  that  the  forces  exerted  by  the  members  cd  and  cb  both  act 
away  from  the  joint,  and,  hence,  the  stress  in  each  of  the  two  members 
will  be  tension  and  thus  we  have  the  stresses  in  the  members  cd  and  cb 
fully  determined. 

Now  as  the  force  exerted  at  c  by  the  member  cb  is  fully  determined, 
the  force  exerted  at  b  by  the  same  member  is  known  and  acts  away  from 
the  joint,  as  indicated.  Then  we  have  only  two  unknown  forces  at  joint  b 
and  hence  the  joint  can  now  be  analyzed.  Then  laying  off  1-3  at  (e), 
equal  and  parallel  to  the  line  1-3  at  (c),  and  3-5  equal  and  parallel  to 
the  line  3-5  at  (c?),  and  drawing  5-6  parallel  to  the  member  bd,  and  1-6 
parallel  to  the  member  be,  we  have  the  force  polygon  1-3-5-6-1  repre- 
senting the  forces  at  joint  b,  where  5-6  and  1-6  represent  the  intensity  of 
the  force  exerted  by  the  member  bd  and  be,  respectively.  Then  by 
scaling  these  lines  we  obtain  the  intensity  of  the  stress  in  the  members 
bd  and  be. 

Now  the  force  exerted  upon  the  joint  b  by  the  member  Ab,  as 
previously  stated,  acts  toward  the  joint  and  the  force  exerted  by  cb  acts 
away  from  the  joint.  Then  transferring  these  directions  to  the  polygon 
at  (e),  it  is  readily  seen  that  the  forces  will  act  around  the  polygon,  as 
indicated,  and  that  the  force  exerted  by  the  member  bd  acts  in  the  polygon 
from  5  to  6,  while  the  force  exerted  by  the  member  be  acts  from  6  to  1, 
and  transferring  these  directions  to  joint  b,  we  have  the  first  acting  away 
from  the  joint  and  the  second  toward  it,  hence  the  stress  in  bd  is  tension 
while  the  stress  in  be  is  compression,  and  thus  we  have  the  stresses  in  the 
two  members  bd  and  be  fully  determined. 

The  force  exerted  at  b  by  the  member  bd  being  determined,  the  force 
exerted  by  the  same  member  at  d  is  fully  known  and  the  same  is  true  ol 
the  force  exerted  at  d  by  the  member  cd.  Then  there  are  but  two  unknown 
forces  at  d  and  hence  the  joint  can  be  analyzed  by  drawing  the  force 
polygon  shown  at  (/)  and  thus  the  stresses  in  the  members  de  and  df  can 
be  determined.  Then  next  the  joints  /  and  e  can  be  analyzed  in  the  same 
way  and  thus  the  stresses  in  all  the  members  in  the  truss  can  be  deter- 
mined, but  one  continuous  diagram  as  shown  at  (/i),  wherein  the  forces  at 
each  joint  and  likewise  the  stress  in  each  member  are  represented,  can  be 
constructed  more  readily  than  the  separate  diagrams,  providing  we  pass 
around  all  the  joints  in  the  same  direction.  Either  direction,  clock- wise 
or  counter  clock- wise,  can  be  taken  for  the  first  joint  analyzed,  but  when 
the  direction  is  once  taken  we  should  pass  around  each  of  the  other  joints 
in  the  same  direction,  for  otherwise  there  is  apt  to  be  confusion. 

For  convenience,  let  S,  £1,  S2,  and  so  on,  represent  the  stresses  in 
the  various  members  of  the  truss,  as  indicated  at  (a). 

To  construct  the  continuous  diagram  at  (&),  let  us  begin  at  joint  A, 
as  before,  and  pass  around  the  joint  counter  clock- wise.  Beginning  with 
the  known  force  R  we  lay  off  1-2  at  (h)  to  any  convenient  scale,  equal 
and  parallel  to  R.  The  next  force  we  come  to  is  the  one  exerted  by  the 
member  Ac.  So  from  2  draw  a  line  %-x  parallel  to  the  member  Ac.  Then 


152  STRUCTURAL  ENGINEERING 

the  next  force  we  come  to  is  the  one  exerted  by  the  member  Ab.  So  from 
1  draw  a  line  parallel  to  this  member,  intersecting  the  line  2-x  at  3,  and 
we  have  the  force  polygon  1-2-3-1  representing  the  forces  at  joint  A, 
where  the  line  2-3  represents  the  intensity  of  the  stress  S  in  the  member 
Ac  and  the  line  1-3  represents  the  intensity  of  the  stress  SI  in  the 
member  Ab.  As  R  acts  upward  the  force  exerted  by  Ac  acts  from  2  to  3 
(in  the  polygon),  and  the  force  exerted  by  the  member  Ab  acts  from 
3  to  1, — all  in  the  same  direction  around  the  polygon.  Transferring  these 
directions  to  joint  A,  we  see  that  S  is  tension  and  S\  compression,  as 
explained  above,  and  thus  we  have  S  and  S~L  fully  determined.  Passing 
on  to  joint  c,  we  have  the  force  exerted  by  the  member  Ac  represented  by 
the  line  2-3  and  acting  from  3  to  2  as  S  is  tension.  The  next  force  we 
come  to,  passing  around  the  joint  counter  clock- wise,  is  PI,  which  we  lay 
off  downward  as  2-4.  Then  from  4  draw  a  line  4-2  parallel  to  the 
member  cd,  and  from  3  draw  a  line  parallel  to  the  member  cb,  and  we 
have  the  force  polygon  3-2-4-5-3,  representing  the  forces  at  joint  c,  where 
the  line  4-5  represents  the  intensity  of  the  stress  S2  in  the  member  cd, 
and  the  line  5-3  represents  the  intensity  of  the  stress  S3  in  the  member 
cb.  As  the  force  exerted  by  the  member  Ac  acts  (in  the  diagram)  from 
3  to  2,  and  PI  from  2  to  4,  the  force  exerted  by  the  member  cd  will  act 
from  4  to  5  and  the  force  exerted  by  the  member  cb  will  act  from  5  to  3. 
Transferring  these  directions  to  joint  c,  we  have  the  force  exerted  by  the 
member  cd  acting  away  from  the  joint  and  the  force  exerted  by  the 
member  cb  acting  away  from  it  also.  Then  S2  and  S3  are  both  tension 
and  thus  we  have  both  fully  determined.  At  j  oint  b  the  force  exerted  by 
the  member  Ab  is  known,  as  its  intensity  was  determined  at  joint  A,  and 
its  direction  of  action  at  b  will  be  in  the  opposite  direction  to  that  of  the 
equal  force  exerted  by  the  same  member  at  A,  and  hence  toward  the  joint 
b,  as  indicated,  and  the  force  exerted  at  b  by  the  member  cd  is  known  as 
its  intensity  was  determined  at  joint  c,  and  its  direction  of  action  will  be 
in  the  opposite  direction  to  that  of  the  equal  force  exerted  by  the  same 
member  at  joint  c,  and  hence  away  from  joint  b.  Then  only  the  forces 
exerted  by  the  members  bd  and  be  remain  to  be  determined  at  joint  b. 
Beginning  with  the  force  exerted  by  the  member  Ab  we  have  its  intensity 
represented  by  the  line  1-3  and  it  acts  from  1  to  3  in  reference  to  joint  b. 
Passing  on  around  the  joint  counter  clock- wise,  we  have  next  the  force 
exerted  by  the  member  cb,  the  intensity  of  which  is  represented  by  the 
line  5-3,  and  it  acts  from  3  to  5.  The  next  force  we  come  to  is  the  one 
exerted  by  the  member  bd.  So  from  5  draw  a  line  5-r  parallel  to  the 
member  bd  and  close  the  polygon  by  drawing  a  line  from  1  parallel  to 
the  member  be,  intersecting  the  line  5-r  at  6,  and  we  have  the  polygon 
1-3-5-6-1  representing  the  forces  at  joint  b.  As  the  force  exerted  by  the 
member  Ab  acts  (in  the  polygon)  from  1  to  3  and  the  one  exerted  by  cb 
acts  from  3  to  5,  the  force  exerted  by  the  member  bd  will  act  from  5  to  6, 
and  the  one  exerted  by  the  member  be  from  6  to  1.  Transferring  these 
directions  to  joint  b  we  have  the  force  exerted  by  bd  acting  away  from 
the  joint  and  the  force  exerted  by  be  acting  toward  it.  Hence  the  stress 
£4  in  bd  is  tension  and  the  stress  S5  in  be\  is  compression,  and  as  the 
intensity  of  £4  is  given  by  the  line  5-6  and  that  of  *S5  by  the  line  6-1,  we 
have  the  stress  in  the  members  bd  and  be  fully  determined. 

Passing  on  now  to  joint  d  we  have  the  forces  exerted  by  the  members 


GRAPHIC  STATICS  153 

bd  and  cd  and  also  the  load  PI  known,  to  determine  the  forces  exerted  by 
the  members  de  and  df.  Beginning  with  the  force  exerted  by  the  member 
bd,  and  passing  around  the  joint  counter  clock- wise,  we  have  that  force 
represented  by  the  line  6-5  and  it  acts  from  6  to  5,  and  the  force  exerted 
by  the  member  cd  represented  by  the  line  5-4  and  it  acts  from  5  to  4. 
Now,  as  P2  acts  downward,  from  4  lay  off  4-7  downward  and  equal  and 
parallel  to  P2  and  from  7  draw  a  line  7-n  parallel  to  the  member  df  and 
close  the  polygon  by  drawing  from  6  a  line  parallel  to  the  member  de 
intersecting  the  line  7-w  at  8,  and  we  have  the  polygon  6-5-4-7-8-6  repre- 
senting the  forces  at  joint  d.  Following  around  this  polygon  in  the 
direction  of  the  action  of  the  known  forces,  that  is,  from  6  to  5,  from 
5  to  4,  on  around  to  6,  we  see  that  the  forces  exerted  by  the  member  df 
and  de  each  act  away  from  joint  d  and  hence  the  stress  in  each  of  the 
members  df  and  de  will  be  tension,  and,  as  the  intensity  of  the  stress  SQ 
in  the  member  df  is  given  by  the  line  7-8  and  the  intensity  of  the  stress 
S7  in  the  member  de  by  the  line  8-6,  we  have  the  stress  in  each  of  the 
members  df  and  de  fully  determined. 

Passing  on  to  joint  /  we  have  the  forces  exerted  by  the  member  df 
and  the  load  P3  known,  to  determine  the  forces  exerted  by  the  members 
fB  and  fe.  The  force  exerted  by  the  member  df  is  represented  by  the 
line  8-7  and  acts  from  8  to  7.  Then  from  7  lay  off  7-9  downward  and 
equal  and  parallel  to  P3  and  drawing  9-10  from  9  parallel  to  the  member 
fB  and  8-10  from  8  parallel  to  the  member  fe,  we  have  the  force  polygon 
8-7-9-10-8  representing  the  forces  at  joint  /  where  the  line  9-10  repre- 
sents the  force  exerted  by  the  member  fB  and  10-8  represents  the  force 
exerted  by  the  member  fe.  Following  around  the  polygon  in  the  direction 
of  the  action  of  the  known  forces,  that  is,  from  8  to  7,  7  to  9,  on  around 
to  8,  we  have  the  force  exerted  by  the  member  fB  and  by  the  member  fe 
both  acting  away  from  the  joint  /,  and  hence  the  stress  SS  in  fB  and  the 
stress  S9  in  fe  are  both  tension  and  as  the  line  9-10  gives  the  intensity  of 
S8,  and  10-8  the  intensity  of  *S9,  we  have  the  stresses  in  the  two  members 
fB  and  fe  fully  determined. 

Now,  passing  on  to  joint  e,  we  have  here  all  of  the  forces  known 
except  the  one  exerted  by  the  member  eB.  Starting  with  the  force 
exerted  by  the  member  be,  and  passing  around  the  joint  counter  clock- 
wise, we  have  the  force  exerted  by  the  member  (be)  given  by  the  line  1-6 
and  it  acts  from  1  to  6 ;  the  force  exerted  by  the  member  de  given  by  the 
line  6-8  and  it  acts  from  6  to  8 ;  and  the  force  exerted  by  the  member  fe. 
given  by  the  line  8-10  and  it  acts  from  8  to  10.  Then  a  line  drawn  from 
10  parallel  to  the  member  eB  should  close  the  polygon  1-6-8-10-1  repre- 
senting the  forces  at  joint  e.  If  this  polygon  should  not  close  it  shows 
that  the  continuous  diagram  has  not  been  accurately  drawn  and  hence 
must  be  wholly  redrawn  to  insure  that  the  intensity  of  the  stresses  thus 
obtained  is  correct.  The  intensity  of  the  stress  £10,  in  the  member  eB, 
is  given  by  the  line  10-1,  and  as  the  force  exerted  by  that  member  at  joint 
e  acts  from  10  to  1  in  the  polygon  1-6-8-10-1,  £10  is  seen  to  be  compres- 
sion and  hence  is  fully  determined,  and  thus  the  stresses  in  all  of  the  truss 
members  are  found. 

For  joint  B  we  have  the  force  polygon  10-9-1-10,  where  the  line  9-1 
represents  the  reaction  Rl,  and  1-10  and  10-9  the  stress  in  eB  and  fB, 
respectively.  The  continuous  diagram  at  (h)  would  be  known  as  a 


154 


STRUCTURAL  ENGINEERING 


"stress  diagram/'  in  fact  all  such  diagrams  representing  stresses  are 
known  as  stress  diagrams. 

It  will  be  observed  that  all  of  the  external  forces,,  which  consist  of 
the  two  reactions  and  the  three  loads,  are  laid  off  on  the  same  line  2-9, 
which  is  known  as  the  load  line;  yet  these  forces,  as  they  hold  the  truss 
in  equilibrium,  really  form  a  closed  polygon,  so  to  speak,  where  the 
reaction  R  is  laid  off  from  1  to  2  and  the  load  PI  from  2  down  to  4,  P2 
from  4  to  7,  P3  from  7  to  9,  and  R~L  from  9  back  to  1 — the  starting  point 
— all  acting,  as  we  may  say,  in  the  same  direction  around  the  polygon 
1-2-4-7-9-1,  which  would  really  be  a  polygon  if  the  forces  were  not 
parallel.  In  constructing  any  stress  diagram  the  force  polygon,  known 
as  the  load  line,  formed  by  the  external  forces  can  be  laid  off  first  and  the 
remainder  of  the  diagram  added  to  this  polygon,  as  shown  in  the  follow- 
ing example: 

Example  2.  Let  it  be  required  to  determine  the  stresses  in  the 
members  of  the  truss  shown  at  (a),  Fig.  127,  due  to  the  three  loads  PI, 
P2,  and  P3.  Let  R  and  Rl  represent  the  reaction  at  A  and  B,  respect- 
ively, due  to  the  three  loads. 


(a) 


Beginning  with  R  and  passing  around  the  truss  as  a  whole  counter 
clock- wise,  and  taking  the  forces  in  consecutive  order,  we  obtain  at  (6) 
the  force  polygon  1-2-3-4-5-1  by  laying  off  1-2  equal  and  parallel  to  R, 
2-3  equal  and  parallel  to  PI,  3-4  equal  and  parallel  to  P2,  4-5  equal  and 
parallel  to  P3,  and  5-1  equal  and  parallel  to  121,  thus  closing  the  polygon. 
As  is  readily  seen,  the  forces  act  in  the  order  1-2-3-4-5-1. 

To  obtain  the  stresses,  we  can  begin  at  either  A  or  B,  say,  B,  and 
passing  around  that  joint  counter  clock- wise  we  obtain  the  polygon 
1-11-5-1,  representing  the  forces  at  that  joint,  by  drawing  1-11  parallel 
to  gB  and  5-11  parallel  to  fB.  The  line  1-11  gives  the  stress  in  gB  and 
5-11  the  stress  in  fB.  As  R\  acts  upward  from  5  to  1,  the  force  exerted 
by  gB  will  act  from  1  to  11,  and  hence  toward  the  joint,  and  the  force 
exerted  by  fB  will  act  from  11  to  5,  and  hence  away  from  the  joint. 
Thus  it  is  seen  that  the  stress  $12,  in  the  member  gB,  is  compression, 
while  the  stress  £10,  in  fB,  is  tension. 

Passing  on  to  joint  /,  we  obtain  the  polygon  4-5-11-10-4  by  drawing 
11-10  parallel  to  gf  and  10-4  parallel  to  fd.  We  really  begin  with  the 
load  P3,  which  acts  from  4  to  5.  Then  the  force  exerted  by  fB  comes 
next,  which  acts  from  5  to  11,  and  next  is  the  force  exerted  by  the 


GRAPHIC  STATICS  155 

member  gf  which  we  can  lay  off  from  11  only  as  a  line  parallel  to  gf, 
and  its  length  is  then  obtained  by  drawing  a  line  from  4  parallel  to  fd, 
thus  closing  the  polygon.  In  a  similar  manner  the  polygon  10-11-1-9-10 
for  joint  g  is  obtained,  and  likewise  the  polygon  9-1-8-9  for  joint  et 
3-4-10-9-8-7-3  for  joint  d,  1-8-7-6-1  for  joint  b,  and  3-7-6-2-3  for  joint  c. 
By  scaling  in  each  case  the  line  representing  the  intensity  of  stress  and 
observing  at  the  same  time  the  direction  of  action  of  the  force  exerted  at 
the  joints,  the  stress  in  each  member  is  fully  determined. 

If  the  direction  around  the  joints  in  examples  1  and  2  be  taken 
clock-wise  instead  of  counter  clock-wise,  as  above,  the  stress  diagram 
would  simply  fall  on  the  left  side  of  the  load  line  instead  of  the  right  as 
shown  at  (k),  Fig.  126,  and  (6),  Fig.  127. 

Stress  diagrams  should  be  drawn  without  the  aid  of  any  letters  or 
marking  of  any  kind  other  than  the  forces  on  the  load  line,  and  the 
stresses  in  the  members  may  be  indicated  in  the  way  shown  above;  how- 
ever, after  a  little  practice,  even  this  marking  will  be  found  to  be 
superfluous.  The  student  should  not  acquire  the  habit  of  laying  off  the 
load  line  and  drawing  the  stress  diagram  without  fully  analyzing  each 
joint  of  the  truss,  and  if  this  is  done,  no  marking  of  the  diagram  is  needed. 

Example  3.  So  far  we  have  considered  loads  only  at  the  bottom 
of  the  trusses,  which  is  often  the  case  for  live  load,  but  for  dead  load, 
especially,  we  have  loads  applied  at  both  the  top  and  bottom  of  the 
trusses  as  shown  at  (a),  Fig.  128.  Let  it  be  required  to  determine  the 
reactions  on  the  truss  shown  there,  and  also  the  stresses  in  the  members, 
due  to  the  six  loads  indicated.  Let  R  and  R1  represent  the  reaction  at 
A  and  B,  respectively,  due  to  these  loads. 

The  most  practical  way  of  determining  the  reactions  is  to  add  the 
top  load  at  each  panel  point  to  the  bottom  load  and  treat  the  sum  as  a 
single  load  in  each  case.  Then  the  ray  diagram  is  constructed  as  shown 
at  (b)  and  an  equilibrium  polygon  as  shown  at  (c)  can  be  drawn.  How- 
ever, the  loads  can  be  laid  off  on  a  load  line  in  any  order  and  a  ray 
diagram  constructed  and  a  corresponding  equilibrium  polygon  drawn. 
For  example,  suppose  the  loads  be  laid  off  on  the  load  line  in  consecutive 
order,  passing  around  the  truss,  as  a  whole,  counter  clock-wise,  as  shown 
at  (d).  Then  taking  G  as  a  pole  and  drawing  the  ray  diagram  at  (d)  and 
beginning  at  any  convenient  point  on  the  line  of  action  of  one  of  the 
forces,  say,  point  1,  on  the  line  of  action  of  PI,  resolve  PI  into  two 
components  /I  and  /2,  which  is  accomplished  by  drawing  /I  (at  point  1) 
parallel  to  the  ray  1-G  and  /2  parallel  to  ray  2-G.  As  fl  and  /2  are 
components  of  PI,  the  first  will  act  in  the  ray  diagram  from  1  to  G  and 
the  second  from  G  to  2,  and  transferring  these  directions  to  point  1  we 
have  them  acting  as  shown.  By  prolonging  the  line  of  action  of  /2  from 
point  1  till  it  intersects  the  line  of  action  of  P2  at  point  2,  and  resolving 
P2  into  two  components  /2  and  /3  by  drawing  /2  parallel  to  the  ray  2-G 
and  /3  parallel  to  the  ray  3-G,  we  have  the  component  at  point  2  balanc- 
ing the  component  /2  at  point  1,  and  drawing  the  line  2-3  parallel  to  the 
ray  3-G,  3-x  parallel  to  the  ray  4-G,  3-4  parallel  to  the  ray  5-G,  and  so 
on,  we  obtain  the  polygon  ?/-l-2-3-,r-3-4-5-.2r,  wherein  the  forces,  or  com- 
ponents, are  all  balanced  except  fl  and  /7.  Then  prolonging  the  line  of 
action  of  fl  until  it  intersects  the  line  of  action  of  #1  at  K  and  prolonging 
the  line  of  action  of  /7  until  it  intersects  the  line  of  action  of  R  at  H,  and 


156 


STRUCTURAL  ENGINEERING 


drawing  the  closing  line  HK,  we  have  the  equilibrium  polygon  K-y-1-2-3- 
x-3-4:-5-H-K  closed,  and  by  drawing  E'-G,  at  (d),  parallel  to  the  closing 
line  HK  we  have  the  reaction  R  given  by  the  line  E'-l  and  the  reaction 
Rl  by  the  line  E'-7.  The  polygon  could  be  closed  just  as  well  by  draw- 
ing z-K'  and  1-H' ',  and  then  H'-K' ,  which  is  the  closing  line  in  that  case, 
and  which  is  parallel  to  H-K. 

As  another  case,  suppose  the  loads  are  laid  off  on  the  load  line,  as 
we  may,  just  in  any  order  as  shown  at  (e).  Then  constructing  the  ray 
diagram  as  shown  we  can  draw  the  equilibrium  polygon  ^-6-7-8-9-10- 
t-W-U-F  as  shown  at  (/),  where  UV  is  the  closing  line.  Then  drawing 


Fig.   128 

E"-P  parallel  to  U-V  we  have  the  reaction  R  given  by  the  line  E"-F  and 
Rl  by  the  line  E"-T. 

Complicated  diagrams  and  polygons  should  be  avoided  as  much  as 
possible.  We  can  accomplish  this  result  only  by  judicious  selection  of 
order  and  position,  for  which  no  fixed  rule  can  be  given,  and  one  must  be 
guided  by  experience  and  common  judgment.  Complicated  diagrams  and 
polygons  are  more  objectionable  on  account  of  inaccuracy,  as  a  rule,  than 
on  account  of  difficult  construction.  For  example,  it  is  obvious  that  there 
is  more  chance  of  error  in  constructing  the  polygon  at  (/)  than  there  is 
in  the  case  of  the  one  at  (c). 

The  most  practical  way  of  obtaining  the  stresses  in  the  truss  shown 
at  (a)  is  to  lay  off  the  load  line  as  shown  at  (#),  which  is  accomplished 


GRAPHIC  STATICS 


157 


by  beginning  with  R  and  passing  around  the  truss  counter  clock-wise, 
taking  the  forces  in  consecutive  order.  Thus  we  lay  off  at  (gr)  the  line 
1-2  upward,,  equal  and  parallel  to  R.  Then  from  2  lay  off  2-3  downward, 
equal  and  parallel  to  PI,  3-4  equal  and  parallel  to  P2,  4-5  equal  and 
parallel  to  P3,  and  from  5  lay  off  5-6  upward,  equal  and  parallel  to  R1, 
and  from  6  lay  off  6-7  downward  equal  and  parallel  to  P4,  7-8  equal  and 
parallel  to  P5,  and  the  line  8-1  closing  the  polygon  1-2-3-4-5-6-7-8-1 
should  be  equal  and  parallel  to  P6.  After  the  load  line  is  laid  off,  we 
can  start  either  from  joint  A  or  B,  and,  passing  around  each  joint  counter 
clock- wise,  the  stress  diagram  shown  at  (#)  can  be  readily  constructed. 
97.  Oblique  Reactions. —  As  a  rule,  the  dead  and  live  load  upon 
structures  act  vertically,  producing  vertical  reactions,  but  the  pressure 
due  to  the  wind,  known  as  wind  load,  produces  reactions  which  are  not 
vertical.  However,  the  method  of  determining  such  reactions  is  prac- 


(b) 


Fig.   129 


tically  the  same  as  for  vertical  reactions.  As  an  example,  let  the  diagram 
at  (a),  Fig.  129,  represent  a  truss  supporting  five  wind  loads,  PI — P5, 
applied  normally  to  the  sloping  side  of  the  truss  as  indicated.  In  case 
the  truss  be  equally  fixed  at  its  supports  A  and  B,  the  reactions  R  and  Rl 
will  be  parallel  to  the  loads  as  indicated,  or  in  case  the  loads  are  not  all 
parallel,  the  reactions  will  be  parallel  to  their  resultant. 

To  determine  the  reactions  R  and  #1,  first  lay  off  the  load  line  CD 
at  (b)  and  draw  the  ray  diagram  as  shown  and  construct  the  equilibrium 
polygon  obcdega  at  (a)  the  same  as  for  any  other  forces.  Then  by 
drawing  OE  in  the  ray  diagram  parallel  to  the  closing  line  ag  of  the 


158 


STRUCTURAL  ENGINEERING 


equilibrium  polygon,  we  have  the  reaction  R  given  by  the  line  EC  and  .Rl 
by  the  line  ED. 

There  is  nothing  about  the  equilibrium  polygon  at  (a)  out  of  the 
ordinary,  except  at  point  a.  At  e  the  force  P5  is  resolved  into  two  com- 
ponents,, and  the  line  of  action  of  the  one  to  the  right  (/o)  is  prolonged 
until  it  intersects  the  line  of  action  of  J?l  at  g.  Now,  at  a  exactly  the 
same  method  of  procedure  is  followed,  but  as  PI  and  R  have  the  same 
line  of  action,  the  segment  parallel  to  the  ray  CO  does  not  appear  as  its 
length  is  zero. 

After  the  reactions  are  known,  the  stress  in  the  individual  members 
of  the  truss  can  be  graphically  determined,  as  explained  above,  by  begin- 
ning at  either  joint  A  or  B. 

In  case  one  end  of  a  truss  be  supported  upon  rollers,  the  reaction  at 
that  end  will  be  vertical,  as  the  rollers  will  not  resist  any  horizontal  force. 
In  such  cases  we  always  know  the  direction  of  the  one  reaction,  but,  as  a 
rule,  the  intensity  of  that  one  and  the  intensity  and  direction  of  the  other 
remain  to  be  determined. 

Let  the  diagram  at  (a),  Fig.  130,  represent  the  same  truss  and  loads 
as  represented  at  (a),  Fig.  129.  First,  suppose  the  truss  supported  upon 


Fig.   130 

rollers  at  B  and  upon  a  fixed  bearing  at  A.  Then,  the  end  B  being  upon 
rollers,  the  reaction  there,  indicated  by  R,  will  act  vertically  and,  conse- 
quently, we  know  its  line  of  action,  while  the  reaction  at  A,  indicated  as 
Rl,  will  evidently  act  obliquely,  as  it  resists  the  horizontal  component  of 
the  wind;  but,  further  than  this,  R\  is  unknown. 

Lay  off  the  load  line  CD  at  (6),  Fig.  130,  and  draw  the  ray  diagram 
the  same  as  in  any  other  case,  and  construct  the  equilibrium  polygon 
AabcdeA  at  (a)  by  beginning  at  A,  the  point  of  application  of  the 
oblique  reaction.  Then  treating  this  equilibrium  polygon  as  a  truss,  we 
can  obtain  R  by  analyzing  joint  e,  as  the  stress  in  the  member  ed  is 
given  by  ray  6  in  the  ray  diagram  at  (b).  Then  by  drawing  from  D,  at 
(k),  the  line  DE  parallel  to  R  and  closing  on  0  with  line  OE  drawn  from 
O  parallel  to  the  member  eA  (at  e),  we  have  the  intensity  of  R  given 
by  the  line  DE.  Then,  as  the  forces  PI — P5  and  the  two  reactions  R 
and  Rl  are  a  system  in  equilibrium,  they  will  form  a  closed  polygon, 


GRAPHIC  STATICS 


159 


B 


DECD,  wherein  the  forces  act  in  the  same  direction  around  the  polygon. 
So,  evidently,  the  line  EC,  at  (6),  represents  the  other  reaction  Rl  in 
intensity  and  direction.  However,  the  analysis  of  joint  A  will  show  this, 
for  beginning  with  PI,  we  have,  at  (6),  the  force  polygon  CmOEC. 

In  case  the  truss  were  supported  upon  rollers  at  A  and  upon  a  fixed 
bearing  at  B,  we  would  begin  at  B,  as  that  would  be  the  point  of  applica- 
tion of  the  oblique  reaction,  and  construct  an  equilibrium  polvgon  as 
Ba'b'c'd'e'H'  and  then  analyze  the  points  Hf  and  B  to  obtain  the  reac- 
tions R2  and  R3,  whence  their  intensity  and  direction  would  be  given,  at 
(6),  by  the  lines  CF  and  FD,  respectively. 

98.  Method  of  Drawing  an  Equilibrium  Polygon  through  Two 
and  Three  Given  Points. —  As  a  case  of  two  points,  let  PI,  P2,  and  P3, 
Fig.  131,  represent  three  given  forces  and  let  A  and  B  be  the  two  given 
points.  Imagine  the  line  AB,  joining  the  two  points,  as  being  a  beam 
supporting  the  three  forces.  Then  the  reactions  on  this  beam,  due  to  the 
three  forces,  will  be  parallel  to  the  resultant  of  the  forces.  By  con- 
structing the  force  polygon  CabDC,  at  (6),  we  have  the  resultant  of  the 
forces  given  by  the  line  CD.  Then  the  reaction  at  A  and  B,  represented, 
respectively,  as  R  and  721,  can 
be  drawn,  as  shown,  parallel  to 
the  line  CD.  Taking  any  point 
O,  at  (&),  as  a  pole  and  con- 
structing the  ray  diagram  as 
shown,  we  can  draw  the  equi-  . 
librium  polygon  A-l-2-3-B'-A, 
as  shown  at  (a).  Then  drawing 
OE  parallel  to  the  closing  line 
AB'  we  have  R  given  by  the 
line  EC  and  Rl  by  the  line  ED. 
These  reactions  will,  of  course, 
be  constant  regardless  of  the 
slope  of  the  closing  line  of  the 
equilibrium  polygon  used  to  de- 
termine them.  So,  regardless  of 
the  location  of  the  pole  in  the 
ray  diagram,  the  line  drawn 
through  it  and  parallel  to  the 
closing  line  of  the  correspond- 
ing equilibrium  polygon  will 

pass  through  E.  Then  evidently  the  pole  of  any  ray  diagram,  where  the 
corresponding  equilibrium  polygon  passes  through  both  A  and  B  will  be 
on  a  line  through  E  parallel  to  AB  as  Ex,  and  as  any  equilibrium  polygon 
drawn  from  A,  having  a  closing  line  parallel  to  AB,  will  pass  through 
both  A  and  B,  it  follows  that  any  point  on  the  line  Ex  can  be  taken  as  a 
pole  and  the  corresponding  equilibrium  polygon  will  pass  through  the  two 
points  A  and  B.  Thus,  taking  01  on  the  line  Ex,  at  (&),  as  a  pole,  and 
constructing  the  ray  diagram,  as  shown,  the  equilibrium  polygon  J -4-5-6-5 
can  be  drawn,  and  taking  O2  as  a  pole  and  constructing  the  ray  diagram 
to  the  left  of  the  load  line,  as  shown,  the  equilibrium  polygon  A-7-8-9-B 
can  be  drawn. 


Fig. 


160 


STRUCTURAL  ENGINEERING 


As  a  case  of  three  points,  let  PI — P5,  shown  at  (a),  Fig.  132,  repre- 
sent five  given  forces  and  let  A,  B,  and  C  be  three  given  points  through 
which  an  equilibrium  polygon  is  to  be  drawn. 

First  construct  the  ray  diagram  at  (b)  by  laying  off  the  forces  in 
consecutive  order,  to  any  convenient  scale,  thus  obtaining  the  load  line 
DabGcF,  and  taking  any  point  0  as  a  pole  and  drawing  the  rays. 

Imagine  the  line  AB,  joining  the  two  points  A  and  B,  as  being  a 
beam  supporting  the  three  forces  PI,  P2,  and  P3  (as  loads).  Then  the 
reactions  on  beam  AB,  at  A  and  B,  due  to  the  three  forces,  will  be 
parallel  to  the  resultant  of  the  three  forces,  which  is  represented  by  the 
line  DG  in  the  ray  diagram.  Then  by  drawing  a  line,  as  yy,  through 
each  of  the  points  A  and  B,  parallel  to  DG,  we  have  the  lines  of  action 
of  the  two  reactions  on  the  beam  AB  which  are  indicated  as  R  and  Rl. 

We  can  consider  the  part  01) G  of  the  ray  diagram  as  pertaining  to 
beam  AB.  By  drawing  the  equilibrium  polygon  A-l-2-3-m-A,  as  shown 


Fig.   132 

at  (a),  and  then  drawing  OE  parallel  to  the  closing  line  Am,  and  from 
E  drawing  a  line  parallel  to  AB,  we  have  the  line  Ex,  any  point  of  which 
could  be  taken  as  a  pole  for  a  ray  diagram,  and  the  corresponding 
equilibrium  polygon  would  pass  through  the  points  A  and  B  as  shown 
above  for  the  case  of  two  points. 

Next,  imagine  the  line  BC  as  being  a  beam  supporting  the  two  forces 
P4  and  P5.  Then  the  reactions  on  this  beam  at  B  and  C  would  be 
parallel  to  GF  (shown  at  (6))  and  by  drawing  the  lines  it  through  B 
and  C  parallel  to  GF  we  have  the  lines  of  action  of  the  reactions  on  the 
beam  BC  which  are  represented  as  R2  and  R3.  The  part  OGF  of  the  ray 
diagram  can  be  considered  as  pertaining  to  the  beam  BC.  Then  drawing 
the  equilibrium  polygon  n-4-5-A-n  and  next  the  line  OEl  parallel  to  the 
closing  line  nh  and  from  El  drawing  a  line  parallel  to  BC  we  have  the 


GRAPHIC  STATICS 

line  ~E\-z,  any  point  of  which  could  be  taken  as  a  pole  for  a  ray  diagram, 
and  the  corresponding  equilibrium  polygon  would  pass  through  the  two 
points  B  and  C,  provided,  of  course,  that  the  polygon  be  started  from 
one  of  the  points.  But,  as  any  point  on  the  line  E-x  can  be  taken  as  a 
pole  for  a  ray  diagram,  and  the  corresponding  equilibrium  polygon  would 
pass  through  points  A  and  B,  provided  it  be  started  from  one  of  the 
points,  undoubtedly  if  the  point  T,  the  intersection  of  E-x  and  El-z,  be 
taken  as  a  pole  of  a  ray  diagram,  the  corresponding  equilibrium  polygon 
will  pass  through  all  three  points  A,  B,  and  C,  if  started  at  one  of  the 
points. 

Problem  1.  Determine  graphically  the  bending  moment  on  the  beam 
AB,  as  shown  in  Fig.  125,  at  load  P3  due  to  the  loads  PI,  P2,  P3,  and  P4, 
assuming  that  each  of  the  loads  weighs  8,000  Ibs.  and  that  all  are  spaced  4 
feet  apart  along  the  beam  and  load  PI  4  feet  from  R,  and  load  P4  5  feet 
from  Rl. 

Problem  2.  Determine  graphically  the  stress  in  the  members  of  the 
truss  shown  in  Fig.  126  due  to  the  loads  indicated.  Assuming: 

PI  =  8,000  Ibs.,  P2  =  10,000  Ibs.,  and  P3  =  12,000  Ibs. 

Length  of  span  =  4  panels  @  25'  =  100'. 

Height  of  truss  =  28'. 

Problem  3.  Determine  graphically  the  stress  in  the  members  of  the 
truss  shown  in  Fig.  127  due  to  the  loads  indicated.  Assuming: 

PI  =  20,000  Ibs.,  P2  =  30,000  Ibs.,  and  P3  =  20,000  Ibs. 

Length  of  span  =  4  panels  @  26'  =  104'. 

Height  at  c  and  /  =  28'. 

Height  atd  =  34'. 

Problem  4*  Determine  graphically  the  stress  in  the  members  of  the 
truss  shown  in  Fig.  128  due  to  the  loads  indicated.  Assuming: 

PI  =  22,000  Ibs.,  P2  =  24,000  Ibs.,  P3  =  25,000  Ibs. 

P4  =  8,000  Ibs.,  P5  =  7,000  Ibs.,  P6  =  9,000  Ibs. 

Length  of  span  =  4  panels  @  26'  =  104'. 

Height  of  truss  =  31'. 


CHAPTER  IX 

INFLUENCE  LINES 

99.  Definition. — An  influence  line  is  a  line  showing  the  intensity 
and  variation  of  reactions,  shears,  moments,  and  stresses,  produced  on 
beams  or  trusses  by  a  single  moving  load.     Owing  to  convenience  the  single 
moving  load  is  taken  as  a  unit  load. 

100.  Influence  Line  for  Reactions  and  Shears  on  a  Simple 
Beam. — Let  AB,  Fig.  133,  represent  a  simple  beam  supporting  a  single 
moving  load  P  and  let  R  and  Rl  represent  the  reaction  at  A   and  B, 
respectively,  due  to  this  load  P  when  at  any  point  on  the  beam. 

Suppose  that  the  load  P  moves  over  the 
beam  from  B  to  A;  the  shear  on  all  vertical 
sections  of  the  beam  between  A  and  the  load  P, 
due  to  that  load,  at  all  times  will  be  equal  to  R, 
which  varies,  of  course,  with  the  position  of 
the  load.  Then  for  the  shear  at  all  points  be- 
tween A  and  the  load,  including  the  point  at 
the  load,  we  have  R  =  Px/L,  which  is  an  equa- 
tion to  a  straight  line. 

Fig.  133  From  the  equation  we  have  R  =  P  when 

x  =  L,  and  R  =  Q  when  ,r  =  0.     Then  if  we  draw 

the  horizontal  line  ab  (=L)  and  erect  the  perpendicular  ac  equal  to  P 
(by  scale)  and  draw  the  line  cb,  any  ordinate  as  y  to  the  line  cb  will  be 
equal  to  the  reaction  at  A  and  also  to  the  shear  on  the  vertical  section  of 
the  beam  just  over  the  ordinate  when  the  load  P  is  at  that  point,  and  from 
this  it  is  seen  that  the  reaction  R  and  the  shear  in  the  beam  vary  from 
0  to  P  as  the  load  moves  from  B  to  A.  So  then  the  line  cb  is  the  influence 
line  for  the  reaction  at  A  and  also  for  the  shear  in  the  beam  when  the 
single  moving  load  moves  from  B  to  A.  In  case  the  load  moved  from  A 
to  B,  the  line  ad  would  be  the  influence  line  for  the  reaction  at  B  and  for 
the  shear  in  the  beam  if  bd  be  laid  off  equal  to  P. 

As  an  example  of  application,  suppose  we  wish  to  determine  the 
reaction  at  A  and  B  and  the  shear  on  the  beam  due  to  a  load  of  20,000 
Ibs.  at  C  and  a  load  of  30,000  Ibs.  at  M.  For  the  sake  of  illustration, 
suppose  P,  the  single  moving  load  referred  to  above  (which  is  really  a 
mythical  load  and  is  not  on  the  beam  at  all  and  is  used  only  to  construct 
the  influence  line),  is  equal  to  1,000  Ibs.  Then  the  ordinate  2  represents 
(by  scale)  what  the  reaction  R  would  be  if  the  1,000-lb.  load  were  at  C. 
Then,  evidently,  the  reaction  at  A,  due  to  the  20,000-lb.  load  at  C,  would 
be  20  times  as  much  or  20s,  and  similarly  the  reaction  at  A  due  to  the 
30,000-lb.  load  at  M  will  be  30s,  and  hence  the  reaction  at  A  due  to  both 
the  20,000-  and  30,000-lb.  loads  is  equal  to  20*  4-  30*,  which  is  also  equal 
to  the  shear  anywhere  between  A  and  C,  while  the  reaction  at  B  and  also 
the  shear  anywhere  between  B  and  M  is  equal  to  20v  +  3Qt  and  the  shear 

162 


INFLUENCE    LINES 


163 


anywhere  between  M  and  C  is  equal  to  20*  +  30*  -  20,000  or  2Qv  +  30*  - 
30*000. 

The  value  of  any  ordinate,  as  z,  s,  etc.,  will  always  be  given  to  the 
same  scale  as  that  used  in  laying  off  the  single  moving  load  P.  It  is 
always  convenient  to  take  this  single  moving  load  P  as  unity,  for  in  that 
case  it  does  not  appear  in  the  computations  at  all. 

Example  1.  Determine  the  maximum  reaction  and  shear  on  a  simple 
beam  30  ft.  long  due  to  the  wheel  loads  as  per  diagram. 

The  placing  of  the  wheels  in 
order  to  obtain  the  maximum  re- 
action or  shear  on  a  simple  beam  is 
really  a  matter  of  trial,  yet  in  most 
cases  we  can  determine  the  position 
by  mere  inspection.  As  in  this  case, 

we  can  readily  see  that  the  maximum  reaction  will  occur  when  wheel  2 
is  at  the  end  of  the  beam  and  wheel  1  off  of  the  beam  altogether.  Then, 
to  obtain  the  maximum  reaction,  draw  the  line  ab  (Fig.  134)  to  represent 
the  length  of  the  beam  to  some  convenient  scale,  say,  -jV'  =  l'-0",  and 
space  the  loads  to  the  same  scale,  placing  wheel  2  at  a.  Then  wheels  2,  3, 
4,  5,  6,  and  7  will  be  on  the  beam  as  shown  in  Fig.  134.  Next  draw  the 
vertical  ordinate  ac  to  represent  1  Ib.  to  a  convenient  scale,  say,  1  in.  = 
1  Ib.,  and  then  draw  the  influence  line  cb.  Next  draw  the  vertical  lines 
through  the  loads  as  shown  and  we  are  then  ready  to  find  the  reaction  at  a, 
which  we  do  in  the  following  manner : 

Take  a  pair  of  dividers,  put  one  of  the  points  at  c  and  bring  the  other 
point  to  a,  then  we  have  the  length  ac  on  our  dividers ;  then  set  one  point 
at  e  and  rest  the  other  point  at  o  (eo  =  ca)  ;  hold  the  point  firmly  at  o  and 

open  the  dividers,  bringing 
the  point  e  to  /;  then  we  will 
have  the  length  of  ca  and  ef 
on  our  dividers.  Then  put 
one  point  of  the  dividers  at  g 
and  rest  the  other  point  at  o' 
(o'g  =  ca  +  ef  —  of)  ;  holding 
the  point  firmly  at  o'  bring 
the  point  at  g  down  to  h. 
Then  put  one  point  of  the 
dividers  at  k  and  rest  the 
other  at  o"  (o"k  =  ac  +  ef  + 
gh)  ;  holding  the  point  firmly 
at  o"  bring  the  point  at  k 
down  to  m,  and  the  distance 
o"m  on  the  dividers  equals 
the  sum  of  the  ordinates  ac, 
ef,  gh,  and  km,  each  of  which 
is  an  ordinate  under  a  20,000- 
Ib.  load.  Then  lay  the  di- 
viders on  a  scale  divided  to  1/10  inches  and  suppose  we  find  that  we  have 
3.03  inches  (=o"m).  Then  the  reaction  at  a  due  to  the  four  20,000-lb. 
loads  is  equal  to  20,000x3.03  =  60,600  Ibs.  Then  with  our  dividers  we 
get,  in  the  same  manner,  the  distance  nt  and  rp,  the  sum  of  which  suppose 


Fig.    134 


164 


STRUCTURAL  ENGINEERING 


we  find  is  0.25  of  an  inch.  Then  the  reaction  at  a  due  to  the  two  13,000- 
Ib.  loads  is  equal  to  13,000  x  0.25  =  3,250  Ibs.  Then  the  total  reaction  at  a 
is  60,600  +  3,250  =  63,850  Ibs.  In  the  case  of  wheel  loads,  as  a  rule,  time 
can  be  saved  by  using  the  ordinates  at  the  centers  of  gravity  of  the  dif- 
ferent groups  of  wheels  instead  of  the  ordinates  directly  under  the  wheels. 
For  example,  the  maximum  reaction  in  the  above  case  is  equal  to 

80,000  x/  +  26,000  x#". 

The  maximum  reaction  at  b  due  to  the  above  wheel  loads  would  be 
the  same  as  at  a,  and  would  be  determined  in  the  same  manner,  but  of 
course  the  loads  and  the  influence  line  would  be  reversed,  end  for  end, 
from  the  position  shown  in  Fig.  134. 

To  determine  the  maximum  shear  at  any  intermediate  point  in  the 
beam,  the  first  thing  to  do  after  constructing  the  influence  line  is  to  place 
some  wheel  which  we  think  will  be  at  the  point  in  question  when  the 
maximum  shear  occurs.  Then  determine  the  end  reaction  in  the  same 
manner  as  was  shown  above  and  subtract  the  intervening  loads,  if  any, 
from  the  reaction,  and  the  difference  will  be  the  shear  at  the  point. 

Thus,  let  ab  (Fig.  135)  represent  a  beam  to  a  convenient  scale,  and 
let  it  be  required  to  find  the  maximum  shear  at  point  d  due  to  a  system  of 
wheel  loads  as  shown.  First  construct  the  influence  line  cb,  making  ac 

L 


cL  O  Oi  (ft  ($7ir?l7rf^ 


Fig.   135 


Fig.   136 


equal  unity.  Say  we  place  wheel  2  at  d,  as  shown,  and  find  the  reaction  at 
a  in  the  same  manner  as  was  shown  above.  Then  subtract  wheel  1  from 
this  reaction  and  the  difference  will  be  the  shear  at  d.  If  there  be  any 
question  as  to  this  being  the  maximum  shear  at  d,  wheels  1  and  3,  and 
possibly  4,  should  each  in  turn  be  placed  at  d  and  the  shear  computed. 
In  this  way  we  can  readily  ascertain  the  wheel  to  place  at  d  for  maximum 
shear  at  that  point. 

101.  Influence  Line  for  Bending  Moments  on  Simple  Beams.— 
As  a  general  case,  let  AB  (Fig.  136)  represent  a  simple  beam  of  length 
L  and  let  cc  be  any  section  a  distance  from  A  and  b  distance  from  B. 
Imagine  a  single  load  P  moving  over  the  beam  from  B  to  A.  The  reaction 
at  A  due  to  this  load  P,  at  any  time,  will  be  R  =  Px/L.  When  P  is  to  the 
right  of  CC)  the  moment  at  that  section  is 


which  is  an  equation  to  a  straight  line  wherein  Mr  —  0  when  x  —  0,  and 
Mr  =  P(b/L)a  when  x  =  b.     Then  drawing  A'B'  (=L)  as  a  reference 


INFLUENCE    LINES 

line,  and  laying  off  om  =  P(b/L)a,  we  can  draw  the  line  mBf,  which 
evidently  is  the  influence  line  for  the  bending  moment  at  the  section  cc 
for  loads  on  the  right  of  the  section,  as  any  ordinate  y  represents  the 
bending  moment  at  the  section  cc  due  to  P  when  P  is  directly  over  that 
ordinate,  and  the  moment  for  any  other  load  can  be  obtained  by  direct 
proportion.  When  P  is  to  the  left  of  cc,  the  moment  at  that  section  is 


which  is  also  an  equation  to  a  straight  line  wherein  Mt  =  0  when  x  —  L  and 
MI  =  P(b/L)a  when  x  —  b.  Then,  evidently,  the  line  mA'  is  the  influence 
line  for  the  bending  moment  at  the  section  cc  for  loads  to  the  left  of  the 
section,  as  any  ordinate  y'  represents  the  bending  moment  at  the  section  cc 
due  to  P  when  P  is  directly  over  that  ordinate,  and  the  moment  for  any 
other  load  can  be  obtained  by  direct  proportion. 

The  two  lines  B'm  and  mA'  combined  would  be  known  as  the 
influence  line,  which  would  be  referred  to  as  influence  line  A'mB'.  By 
prolonging  the  line  mB'  until  it  intersects  the  line  of  action  of  R  at  n,  we 
have  two  similar  triangles,  A'B'n  and  oB'm.  From  these  similar  triangles 
we  have 

A'n     A'B' 
om        oB' 

or 

A'n       L 
~Ib~  =b 


from  which  we  obtain  A'n  —  Pa.  But  if  P  —  unity,  we  would  have  A'n  —  a, 
and  the  influence  line  for  unit  load  could  then  be  constructed  by  making 
A'n  =  a,  drawing  B'n,  dropping  the  perpendicular  om,  and  drawing  A'm. 
Or  the  same  thing  could  be  accomplished  by  laying  off  B'r  =  b,  drawing 
A'r,  dropping  the  perpendicular  om,  and  drawing  mBf. 

Example  1.  Let  it  be  required  to  determine  the  maximum  bending 
moment  on  a  40-ft.  girder  due  to  the  loading  given  in  Example  1  of  the 
preceding  article.  Wheels  1  to  6  are  the  heaviest  that  can  be  placed  on 
the  girder,  and  consequently  will  very  likely  produce  the  maximum 
moment — this  much  being  determined  by  mere  inspection.  It  is  first 
necessary  to  determine  the  center  of  gravity  of  these  wheels  in  order  to 
place  them  on  the  girder  so  as  to  satisfy  the  position  for  maximum 
moment.  (See  Art.  88.)  In  such  problems  it  is  quite  convenient  to 
determine  the  center  of  gravity  by  means  of  proportional  triangles.  (See 
Art.  45.)  This  is  accomplished  in  the  following  manner: 

First  lay  out  the  diagram  of  the  loads  to  a  convenient  scale  (say, 
y  —  l'-0")  as  shown  on  line  AB  in  Fig.  137.  We  know  from  observation 
that  the  center  of  gravity  of  wheels  2  to  5  inclusive  is  at  d.  Then  the 
center  of  gravity  of  wheels  2  to  5  inclusive  being  at  d,  the  problem  reduces 
to  finding  the  center  of  gravity  of  wheels  1,  6,  and  the  80,000  pounds 
acting  through  d.  From  wheel  1  draw  the  line  ax,  making  a  convenient 
angle  with  line  AB  and  lay  off  ab  =  80,000  Ibs.  and  be  =  10,000  Ibs. 
(weight  of  wheel  1)  to  any  convenient  scale.  Then  join  c  and  d  and 


166 


STRUCTURAL  ENGINEERING 


through  b  draw  a  line  parallel  to  cd  and  the  point  e  where  it  intersects 
the  line  AB  will  be  the  center  of  gravity  of  all  of  the  wheels  from  1  to  5 
inclusive.  Next,  from  e  draw  the  line  exf  at  random,  making  any  con- 
venient angle  with  AB,  and  lay  off  ef  =  13,000  Ibs.  (weight  of  wheel  6) 


Fig.   137 

and  /<7  =  90,000  Ibs.  (weight  of  wheels  1  to  5).  Then  join-gr  and  h,  and 
through  /  draw  a  line  parallel  to  gh  and  the  point  k  where  it  intersects  the 
line  AB  is  the  center  of  gravity  of  all  of  the  wheels  from  1  to  6  inclusive, 
which  was  desired. 

Now,  as  this  point  k  is  nearest  wheel  4,  the  maximum  moment  will 
(very  likely)  occur  under  that  wheel,  and  the  line  SS  bisecting  the  dis- 
tance from  k  to  wheel  4  will  be  the  center  of  the  span.  Then,  by  laying 

off  20  ft.  on  each  side  of  this  line 
SS  to  the  same  scale  as  the  load 
diagram,  we  have  the  girder  in 
position  represented  as  DE.  By 
dropping  the  perpendicular  from 
wheel  4  we  have  the  point  o  on  the 
beam  where  the  maximum  moment 
occurs.  The  influence  line  DmE 
for  the  moment  at  that  point  is  then 
constructed  by  laying  off  EF  =  oE 
and  drawing  FD,  om,  and  mE,  as 
explained  above.  Then  by  drop- 
ping perpendiculars  down  from  the  loads,  we  have  the  ordinates  1,  2,  3, 
...  6,  and  by  multiplying  each  by  the  load  above  it,  and  adding  these 
products,  we  will  have  the  bending  moment  at  o,  which  is  the  maximum 
bending  moment  on  the  beam.  The  sum  of  the  ordinates  2,  3,  4,  and  5 


Fig.   138 


INFLUENCE    LINES  167 

can  be  obtained  by  the  use  of  dividers  as  explained  in  the  preceding 
article. 

Example  2.  As  a  case  of  uniform  load,  let  AB  (Fig.  138)  represent 
a  simple  beam  supporting  a  uniform  load  of  p  pounds  per  foot,  and  let  o 
be  any  section  a  distance  from  A  and  b  distance  from  B.  The  influence 
line  AmB  for  the  bending  moment  at  o  is  constructed  in  the  same  manner 
as  explained  above  by  laying  off  An  =  Ao  =  a,  then  drawing  nB,  om,  and 
Am  in  consecutive  order.  From  proportional  triangles  we  obtain 

b 

om  —  —  a. 
L-J 

We  can  consider  the  uniform  load  as  made  up  of  infinitesimal  loads 
each  weighing  pdx  pounds.  Then  for  the  bending  moment  at  o  due  to  any 
such  infinitesimal  load  at  x  distance  to  the  right  of  o  we  have 

dM  =  pdxy (1). 

But  from  proportional  triangles  we  have 

y        _b-x 


(I) 


from  which  we  obtain 


Now  substituting  this  value  of  y  in  (1),  we  have 

dM  =  £  a(b  -x^dx  ....................................  (2). 

Li 

Then  for  the  bending  moment  at  o  due  to  all  such  loads  to  the  right,  we 
have 


But  ^(b/L}axb  =  area  of  the  triangle  omB,  since  the  ordinate  om  — 
(b/L)a.  So  we  have  the  bending  moment  at  o  due  to  the  uniform  load  to 
the  right  equal  to  the  area  of  the  triangle  omB  multiplied  by  the  uniform 
load  per  linear  foot  of  span.  In  a  similar  manner  it  can  be  shown  that 
the  bending  moment  at  o  due  to  the  uniform  load  to  the  left  of  o  is  equal 
to  the  area  of  the  triangle  omA  multiplied  by  the  uniform  load  per  linear 
foot.  Then  letting  M  represent  the  total  bending  moment  at  o  due  to  the 
total  uniform  load  on  the  girder,  we  have 


=  area  of  triangle  AmB  x  p  ......  (3), 

that  is,  the  total  bending  moment  at  o  is  equal  to  the  area  of  the  triangU 
AmB  (formed  by  the  influence  line  and  the  reference  line)  multiplied  by 
the  uniform  load  per  linear  foot  of  span. 


168 


STRUCTURAL  ENGINEERING 


If  the  section  be  taken  at  the  center  of  the  span,  we  have  b  =  a  =  L/2. 
Substituting  in  equation  (3),  we  have 

L     L 


which  is  readily  recognized  as  the  formula  for  the  bending  moment  at  the 
center  of  a  beam  uniformly  loaded. 

102.  Influence  Lines  for  Shears  and  Bending-  Moments  on 
Trusses. — Let  the  diagram  at  (a),  Fig.  139,  represent  a  simple  truss  and 
let  it  be.  required  to  construct  an  influence  line  for  the  shear  in  any  panel 
as  CD.  It  is  evident  that  any  load  as  W  when  at  D  or  to  the  right  of  D 
will  produce  tension  in  the  diagonal  UD,  while  any  load  as  W\  when  at 
C  or  to  the  left  of  C  will  produce  compression  in  the  same  diagonal. 


Then  evidently  there  must  be  some  point  between  C  and  D  where,  if  a 
load  were  placed,  the  stress  in  the  diagonal  due  to  the  load  would  be  zero. 
Then  evidently  the  influence  line  for  the  shear  in  the  panel  will  be  of  the 
form  EdeF  shown  at  (6),  which  is  readily  constructed  for  a  unit  load  by 
laying  off  En  and  Fn'  each  equal  to  unity  and  dropping  the  perpen- 
diculars ef  and  hd  under  the  panel  points  and  drawing  ed.  The  shear 
in  the  panel  producing  tension  in  the  diagonal  UD,  due  to  any  load  as  W 
when  at  any  point  x  distance  from  B,  is  equal  to  Wylf  yl  being  the 
ordinate  under  the  load  as  shown,  while  the  shear  in  the  panel  producing 
compression  in  the  diagonal  due  to  any  load  as  W\  is  equal  to  W\y2. 

It  is  evident  then  that  to  obtain  the  maximum  shear  in  panel  CD 
producing  tension  in  the  diagonal  UD,  which  we  will  call  positive  shear, 
there  should  be  no  loads  to  the  left  of  0,  and  to  obtain  the  maximum 
shear  producing  compression  in  the  same  diagonal,  which  we  will  call 
negative  shear,  there  should  be  no  loads  to  the  right  of  0.  That  is,  one 
kind  of  shear  will  be  a  maximum  when  the  loads  extend  from  B  to  0,  and 
the  other  kind  will  be  a  maximum  when  the  loads  extend  from  A  to  O. 


INFLUENCE    LINES  169 

In  case  of  a  uniform  live  load,  the  maximum  positive  shear  would  be  equal 
to  the  area  of  the  triangle  Fke  multiplied  by  the  load  per  foot,  while  the 
maximum  negative  shear  would  be  equal  to  the  area  of  the  triangle  Edk 
multiplied  by  the  load  per  foot.  In  case  of  uniform  dead  load,  the  shear 
is  equal  to  the  difference  of  the  areas  of  the  two  triangles  Edk  and  Fke 
multiplied  by  the  dead  load  per  foot.  In  the  case  of  concentrated  live 
loads,  as  wheel  loads,  to  obtain  the  maximum  shear  in  the  panel  the  loads 
would  be  so  placed  that  a  load  would  be  at  the  panel  point  and  the  front 
load  as  near  the  point  0  as  possible  (not  beyond).  For  example,  to  obtain 
the  maximum  positive  shear  in  panel  CD  a  load  would  be  placed  at  D,  the 
one  that  would  bring  the  front  load  closest  to  the  point  0,  and  to  obtain 
the  maximum  negative  shear  the  load  would  be  placed  at  C  that  would 
bring  the  front  load  as  near  0  as  possible.  The  same  result  is  obtained 
in  this  manner  (graphically)  as  would  be  obtained  by  satisfying  the 
criterion  of  Art.  90. 

The  influence  line  for  the  shear  in  each  of  the  other  panels  can  be 
drawn  on  the  same  preliminary  construction  at  (&)  as  used  for  panel  CD. 
For  example,  the  influence  lines  for  panel  MC,  LN,  NB  are  Ed'e'F, 
Ed"e"F,  and  Ed"  '  F,  respectively.  The  construction  of  each  is  accom- 
plished in  the  same  manner  as  explained  above  in  the  case  of  panel  CD. 

In  the  case  of  trusses,  the  bending  moments  are  desired  only  at  the 
panel  points.  The  influence  lines  for  the  moments  at  these  points  are 
constructed  the  same  as  though  the  points  were  on  a  simple  beam,  which 
was  treated  in  the  preceding  article.  For  example,  the  influence  line  for 
bending  moment  at  the  panel  point  C  of  the  truss  represented  at  (a) 
(same  for  point  U)  is  constructed  as  shown  at  (c)  by  laying  off  Gs  =  a, 
the  distance  of  the  panel  point  from  A,  and  drawing  sH,  mm' ,  and  mG 
in  consecutive  order,  as  explained  in  the  last  article  for  simple  beams. 
The  influence  line  for  the  bending  moment  at  any  other  panel  point  would 
be  constructed  in  the  same  manner.  The  moment  at  any  panel  point 
would  be  obtained  in  the  same  way  as  explained  in  the  preceding  article 
for  the  case  of  any  point  on  a  simple  beam.  The  exact  placing  of  concen- 
trated loads  for  maximum  moments  at  the  different  panel  points  would  be 
according  to  Art.  91. 

103.  Influence  Lines  for  Stresses  in  Truss  Members. — Instead 
of  constructing  influence  lines  for  the  shears  and  moments  as  in  the  last 
article  and  then  determining  the  stresses  in  the  truss  members  from  these, 
it  is  more  convenient  to  construct  influence  lines  for  the  stresses  in  the 
members,  thereby  obtaining  the  maximum  stresses  directly  from  the 
influence  lines  without  dealing  with  either  shears  or  moments. 

Let  the  diagram  at  (a),  Fig.  140,  represent  a  truss  and  let  it  be 
required  to  construct  an  influence  line  for  the  stress  in  the  diagonal  ED. 
First  draw  the  line  A'B'  at  (b)  (=L)  and  lay  off  A'n  and  B'n'  each  equal 
to  unity  and  draw  nB'  and  A'n'.  Then  by  drawing  the  ordinates  de  and 
fb  and  the  line  db  we  have  the  influence  line  A'bdB'  for  the  shear  in  the 
panel  CD  the  same  as  in  the  preceding  article.  The  ordinate  ed  is  equal 
to  the  positive  shear  that  would  be  produced  in  the  panel  by  a  unit  load 
at  D,  and  the  ordinate  fb  the  negative  shear  in  the  panel  that  would  be 
produced  by  a  unit  load  at  C.  Now,  as  the  stress  in  the  diagonal  ED 
varies  as  the  shear  in  the  panel,  it  is  evident  that  if  the  ordinates  de  and 
bf  were  laid  off  to  represent  the  stress  in  the  diagonal  that  would  be 


170 


STRUCTURAL  ENGINEERING 


produced  by  a  unit  load  at  these  same  points  and  an  influence  line  be 
constructed  accordingly,  we  would  have  an  influence  line  for  the  stress  in 
the  diagonal. 

For  example,  if  e'd' ',  at  (c),  were  equal  to  the  stress  produced  in  the 
diagonal  ED  by  a  unit  load  at  D,  and  f'b'  were  equal  to  the  stress  pro- 
duced in  the  same  by  a  unit  load  at  C,  the  line  A"b'd'B"  would  be  the 
influence  line  for  the  stress  in  the  diagonal.  And  likewise,  if  ft",  at  (c), 
and  r"e'  were  equal,  respectively,  to  the  stress  produced  in  the  post  EC 
by  a  unit  load  at  C  and  D,  the  line  A"t"r"B"  would  be  the  influence  line 
for  the  stress  in  the  post. 

Laying  off  B'k  at  (6),  equal  to 
BD  and  drawing  kA',  ea,  and  aB'  in 
consecutive  order,  we  have  the  influ- 
ence line  for  the  bending  moment  at 
panel  point  D,  as  explained  in  the 
last  article.  Now,  as  the  stress  in 
the  top  chord  EF  varies  directly  as 
this  moment,  it  is  evident  that  if  ev, 
at  (6),  were  equal  to  the  stress  pro- 
duced in  the  chord  by  a  unit  load  at 
D,  the  line  A'vB'  would  be  the  influ- 
ence line  for  the  stress  in  the  chord 
EF.  From  the  above  it  is  evident 
that  an  influence  line  for  the  stress  in 
any  truss  member  can  be  drawn  as 
readily  as  the  influence  lines  for  the 
moments  and  shears  on  the  truss  by 
computing  the  maximum  stress  in  the 
member  considered,  due  to  a  unit 
load  and  using  this  value  as  the  ordi- 
nate  in  establishing  the  influence  line. 

The  stresses  due  to  the  unit  load  can  be  determined  most  readily  by 
graphics. 

As  an  example,  let  us  first  construct  the  influence  line  for  the  stress 
in  the  top  chord  EF.  The  first  thing  to  do  is  to  determine  the  stress  in 
the  member  due  to  a  unit  load  at  D.  Imagine  OO'  (drawn  through  the 
top  chord)  to  be  a  beam,  and  each  of  the  lines  OD  and  DO'  to  be  a  rod 
or  a  rope.  Then  we  would  have  a  structure  OO'D  such  that  the  stress  in 
the  member  00',  produced  by  a  unit  load  at  D,  is  the  same  as  the  stress 
produced  in  the  top  chord  EF  by  a  unit  load  at  the  same  point.  Now  this 
structure  OO'D  is  very  readily  analyzed  graphically.  The  reaction  at  0 
due  to  a  unit  load  at  D  is  given  by  the  ordinate  de,  at  (6).  Then  con- 
sidering the  point  0  and  drawing  from  d  a  line  parallel  to  00'  and  from 
e  a  line  parallel  to  OD  intersecting  the  first  line  at  r,  we  have  the  stress 
in  00'  given  by  the  line  dr  which  is  also  the  stress  in  the  top  chord  EF. 
Then  from  e  lay  off  ev  equal  to  dr,  just  found,  and  the  influence  line 
B'vA'  for  the  stress  in  the  top  chord  EF  is  drawn. 

Next,  let  us  construct  the  influence  line  for  the  stress  in  the  diagonal 
ED.  The  stress  in  the  diagonal  corresponding  to  positive  shear  due  to  a 
unit  load  at  D  is  given  by  the  line  rs,  at  (6),  which  is  drawn  from  r 
parallel  to  the  diagonal,  and  as  bf  is  equal  to  the  negative  shear  in  the 


Fig.   140 


INFLUENCE    LINES  171 

panel  that  would  be  produced  by  a  unit  load  at  C,  by  drawing  the  lines 
OC  and  CO'  and  br'  and  r'f  parallel,  respectively,  to  00'  and  CO',  and 
r's'  parallel  to  the  diagonal,  we  will  have  the  stress  in  the  diagonal  that 
would  be  produced  by  a  unit  load  at  C  given  by  this  last  line  r's'.  Then 
by  laying  off  e'd'  =  rs,  at  (c),  and  fb'  =  r's',  the  influence  line  A"b'd'B" 
is  readily  constructed.  In  like  manner  the  influence  line  A"t"r"B"  for 
the  stress  in  the  post  EC  is  readily  constructed  by  laying  off  the  ordinates 
e'r"  =  rt  (given  at  (6))  and  t"f  =  r'f,  as  it  will  be  readily  seen  that  rt 
is  equal  to  the  stress  in  the  post  due  to  a  unit  load  at  D,  and  r't'  is  equal 
to  the  stress  in  the  same  due  to  a  unit  load  at  C.  The  influence  line  for 
the  stress  in  any  member  of  any  truss  can  be  readily  constructed  in  the 
same  manner  as  shown  for  the  above  cases. 

There  are  other  methods  employed  to  determine  the  stresses  in  the 
truss  members  due  to  the  unit  load,  and  these  will  be  presented  later  as 
the  practical  application  of  influence  lines  is  taken  up. 


CHAPTER  X 

DESIGN   OF    I-BEAMS   AND    PLATE    GIRDERS 

104.  General  Data.— 

Specifications,,  A.  R.  E.  Ass'n.*     (For  steel  railroad  bridges.) 
Allowable  bending  stress  on  beam  and  girder  flanges .  16,000  Ibs.  per  sq.  in. 

Allowable  shearing  stress  on  shop  rivets 12,000  Ibs.  per  sq.  in. 

Allowable  shearing  stress  on  field  rivets  and  webs. .  .  .  10,000  Ibs.  per  sq.  in. 

Allowable  bearing  stress  on  shop  rivets 24,000  Ibs.  per  sq.  in. 

Allowable  bearing  stress  on  field  rivets 20,000  Ibs.  per  sq.  in. 

Allowable  bearing  on  masonry 600  Ibs.  per  sq.  in. 

DESIGN  OF   I-BEAMS 

105.  Outline   of   Usual   Method    of   Procedure. — In  designing 
I-beams  the  first  thing  to  do  is  to  determine  the  maximum  bending  mo- 
ment in  inch  pounds.     Then,  using  Formula  C,  Art.   53,  we  have  the 
bending  moment 

M=tL; 

y 

dividing  through  by  /,  we  have 

M     7 

/  '  y 

The  quantity  I/y  is  known  as  the  "Section  Modulus/'  which  is  given  in 
Tables  1  and  2  in  the  back  of  this  book,  for  practically  all  I-beams,  and 
the  same  will  be  found  in  structural  handbooks,  such  as  Carnegie, 
Cambria,  etc.  So  the  size  of  an  I-beam  required  to  resist  a  known 
bending  moment  can  be  determined  by  simply  dividing  the  bending 
moment  (in  inch  pounds)  by  the  allowable  stress,  which  is  specified  above 
as  16,000  Ibs.,  thus  obtaining  the  required  section  modulus,  and  from  the 
tables  we  find  the  lightest  I-beam  having  this  or  approximately  this 
modulus  and  that  will  be  the  beam  to  use. 

For  example,  suppose  the  bending  moment  for  a  given  span  is 
433,000  inch  pounds.  Then  the  section  modulus  of  the  beam  required 
for  the  span  is 

433,000 
16.000 


Glancing  over  column  10  of  Table  1  we  find  that  a  10-inch  by  30-pound 
beam,  which  has  a  section  modulus  of  26.8,  is  the  nearest  to  the  required 
beam,  and  hence  would  be  used. 

After  an  I-beam  is  designed  to  resist  the  maximum  bending  moment, 
it  can  be  tested  for  shear,  by  dividing  the  maximum  shear  by  the  area  of 

*  Those  specifications  can  be  obtained  from  the  Secretary.  American  Railway  En- 
gineering Association.  900  South  Michigan  Avenue  Chicago,  111.  Twenty-five  cents  per 
copy,  or  less  if  several  copies  are  ordered  at  one  time. 

172 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDEHS 


173 


the  cross-section  of  the  beam.  If  this  should  exceed  10,000  Ibs.  per 
square  inch.,  the  permissible  intensity,  a  heavier  beam  should  be  used  or 
the  web  of  the  beam  should  be  reinforced.  However,  there  are  but  very 
few  cases  where  the  shear  affects  the  design  of  an  I-beam,  owing  to  the 
webs  of  I-beams  being  comparatively  thick. 

An  I-beam  can  be  designed  by  using  Formula  D, 

My 

T~    I   ' 

and  finding  values  for  y  and  I  that  will  give  /  =  16,000  (approximately). 
But,  as  is  readily  seen,  this  is  not  a  very  convenient  method  of  procedure. 
106.     Example  1. — Design  an  I-beam  of   15-ft.   span  to  support  a 
total  uniform  load  of  800  Ibs.  per  foot. 

The    maximum    bending    moment    (=pL2/8)   =  |x800x!52x!2  = 
270,000  inch  Ibs. 


Then, 


270,000 


=  16.8  =section  modulus. 


16,000 

The  I-beam  having  a  section  modulus  nearest  this  value  is  the  8-in. 
by  25.5-lb.,  but  the  9-in.  by  21-lb.  would  be  used  as  it  is  lighter.  It  is 
seen  that  the  8-in.  beams  are  all  too  small  except  the  25.5-lb.  beam. 

107.      Example  2. — Determine  the   size   of   I-beam   required  in  the 
case  of  the  simple  beam  shown  in  Fig.   141,  where  AB  represents  the 


Al 


1 

1 

*l 

J 

W////////A 

'////////////A 

'(//fido^ffii 

&//////////M 

t 


,3'-  0 


Rl 


Fig.   141 

beam  supporting  the  loads  indicated.  By  taking  moments  about  B  we 
find  12  =  19,630  Ibs.  Adding  up  the  forces,  beginning  at  A,  we  find  that 
the  shear  passes  through  zero  at  the  2,000-lb.  load,  hence  the  maximum 
bending  moment  occurs  at  that  load.  (See  Art.  88.)  Then  for  the 
maximum  bending  moment  we  have 

M=[19,630x6-(l,600x6x3)-(9,000x3)]12  =  743,760  inch  Ibs. 

Then,  743,760 

=46.4  =  Section  Modulus. 

1  o,0  00 

So  a  15-in.  by  42-lb.  I-beam  would  be  used.  The  12-in.  by  40-lb.  is  too 
light  and  12-in.  by  45-lb.  is  heavier  than  the  15-in.  by  42-lb.  beam. 

108.  Example  3. — Determine  the  size  of  I-beam  required  in  the 
case  of  the  simple  beam  shown  in  Fig.  142,  where  AB  represents  the 
beam  supporting  a  load  which  varies  from  0  at  A  to  2,000  Ibs.  at  B,  as 
indicated. 

For  convenience,  let  p  (=2,000  Ibs.)  represent  the  intensity  of  the 


174 


STRUCTURAL  ENGINEERING 


load  at  B.     Then  for  the  total  load  on  the  beam  (neglecting  the  weight 
of  the  beam)  we  have 


Fig.    142 


and  for  the  intensity  of  the  load  at  any  point  a;  distance  from  A  we  have 

pz 
L  ' 


Now,  taking  moments  about  B  we  have 

W- 

~          3       W     pL 

~=r:~~ 


for  the  reaction  at  A. 

Now  for  the  bending  moment  at  any  point  x  distance  from  A,  we  have 


When  this  moment  is  a  maximum 
dM     pL     px2 

=~~  


(which  is  obtained  by  differentiating  (1)).     This,  as  is  readily  seen,  will 
occur  when  L2  =  3x2,  from  which  we  obtain 


'  =  LAJ  — =0.58L  (approximately). 


So  the  point  of  maximum  bending  moment  is  0.58Z,  from  A.  Then 
substituting  0.58L  for  x  in  equation  (1)  we  have  for  the  maximum 
bending  moment 

M=  (-  J0.58L2-  (- j0.19L2  =  ^  (0.39L2), 

and  substituting  the  numerical  value  of  L  and  p,  and  multiplying  by  12, 
to  reduce  the  moment  to  inch  pounds,  we  have 


2,000 


(0.39  x  144  x  12)=  224,600  inch  Ibs. 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDERS  175 

for  the  maximum  bending  moment. 

Then,  -—7^——=  14  =  section  modulus. 

lo,UUU 

So  an  8"xl8#  I-beam  would  be  used. 

In  case  the  weight  of  the  beam  were  considered,  we  would  simply 
include  the  moment  of  this  weight  in  an  equation  for  the  bending  moment, 
corresponding  to  (1).  Thus,  if  the  weight  of  the  beam  be  w  pounds  per 
foot,  the  equation  for  the  bending  moment  at  any  point  x  distance  from  A 
would  then  be 

__     pL        px*      wL         tvxz 

M=--+*-—  .........................  3- 


Then  by  placing  dM/dz  =  Q,  the  point  of  maximum  moment  can  be 
determined  and  then  the  maximum  bending  moment  at  that  point  due  to 
the  combined  loads  can  be  obtained  and  the  beam  designed  accordingly. 

109.  Example  4.  —  Determine  the  size  of  I-beam  required  in  the 
case  of  the  overhanging  beam  shown  in  Fig.  143,  where  ABC  represents 
the  beam  supporting  the  loads  indicated. 


Fig.  143 


In  this  case  there  are  two  bending  moments  to  consider;  one  at  B 
and  the  other  at  the  point  of  zero  shear  in  the  span  BC. 
For  the  bending  moment  at  B  we  have 

M=  (4,000x3 +  600x5x2.5)  12  =  234,000  inch  Ibs. 
Then  taking  moments  about  C  we  have 

R  =  ~ (600x15x7.5  +  8,000x4  +  4,000x13)=  15,150  Ibs. 

for  the  reaction  at  B.    Then  for  the  reaction  at  C  we  have  the  total  load 
on  the  beam  minus  R,  that  is, 

Rl  =  21,000  -15,150  =  +5,850  Ibs. 

Now  beginning  at  C  and  adding  up  the  forces  toward  the  left  we 
find  that  the  shear  changes  signs,  that  is,  passes  through  zero,  at  the 
8,000-lb.  load.  So  the  maximum  bending  moment  in  span  BC  occurs  at 
that  load.  Then  taking  moments  about  that  load  and  considering  the 
forces  to  the  right  of  it,  we  have 

M'=(5,850x4-  600x4x2)  12  =  223,200  inch  Ibs. 

for  the  maximum  bending  moment  in  span  BC.     The  bending  moment  at 
B  is  the  greater  and  hence  the  I-beam  must  be  designed  for  that  moment. 


176 


STRUCTURAL  ENGINEERING 


So,  using  the  moment  at  B,  we  have 

234,000 

=  14.0  =  section  modulus, 

-L  O  .U  \J  \) 

and  hence  an  8-in.  by  18-lb.  I-beam  would  be  used. 

110.  Example  5. — Determine  the  size  of  I-beam  required  in  the 
case  of  the  continuous  beam  shown  in  Fig.  144,  where  ABC  represents  the 
beam  supporting  4  ft.  of  uniform  load  in  span  AB  and  a  single  load  of 
18,000  Ibs.  in  span  BC. 


DC  <?=kl)    . 


vc 


ikL,  =  c) 


\PI 


W 


m 


4 


Fig.   144 

Applying  the  three-moment  equation,  (L), 


given  in  Art.  70,  we  have  both  M'  and  M"  '  —  0,  and  the  quantity 


and  the  quantity 

-P'L  l(2k1-3kl 

Then   substituting  these   values   in  the  general  three-moment   equation, 
we  have 


Now  everything  in  this  equation  is  known  except  M",  the  bending  moment 
at  support  B.  Then  by  substituting  the  numerical  values  of  the  known 
quantities,  as  given  in  Fig.  144,  and  reducing,  we  have 

M"=-25,363  ft.  Ibs.  =-304,356  inch  Ibs. 

for  the  bending  moment  at  B. 

Then  taking  moments  about  B,  we  have 

-25,363  =  10121  -  4,000  x  4, 
from  which  we  obtain 

721  =-936  Ibs. 

for  the  reaction  at  A,  and  taking  moments  about  B  and  considering  span 
BC,  we  have 

-  25,363  =  12  R3  -  18,000  x  6, 


DESIGN    OF   I-BEAMS    AND    PLATE    GIRDEES  17r, 

from  which  we  obtain 

R3=  +6,884  Ibs. 

for  the  reaction  at  C. 

Now,  as  the  three  reactions  must  be  equal  to  the  total  load  on  the 
beam,  we  have 

22,000  =-936  +  #2  +  6,886, 
from  which  we  obtain 

£2  =+16,050  Ibs. 

for  the  reaction  at  B. 

Now,  as  all  of  the  reactions  are  determined,  the  moments  and  stresses 
on  the  above  beam  can  be  determined  as  readily  as  for  any  beam.  There 
are  three  bending  moments  to  consider  in  the  above  case ;  the  maximum  in 
each  of  the  spans  and  the  one  at  support  B. 

The  reaction  Rl  being  minus,  it  is  readily  seen  that  the  maximum 
bending  moment  in  span  AB  will  occur  at  support  B,  and  as  span  BC 
supports  only  the  one  load  it  is  evident  that  the  maximum  bending 
moment  in  that  span  will  occur  under  the  load.  Then  we  really  have  only 
the  moment  under  the  18,000-lb.  load  yet  to  determine  as  the  moment  at 
B  is  determined  above. 

Then,  taking  moments  about  the  18,000-lb.  load,  we  have 

A^  =  6x^3  =  6x6,884  =  41,316    foot    Ibs.  =  41,316x12  =  495,792    inch    Ibs. 

for  the  bending  moment  at  that  load,  which  is  greater  than  the  moment  at 
B  and  hence  is  the  maximum  bending  moment  on  the  beam,  and  conse- 
quently the  beam  must  be  designed  to  resist  this  moment.  Thus  we  have 

495,79.2 

=3 1.0  =  section  modulus. 

16,000 

So  a  12-in.  by  31.5-lb.  I-beam  would  be  used. 

DESIGN  OF  PLATE  GIRDERS 

111.  Description. — A  plate  girder  is,  for  the  most  part,  a  built 
I-beam  composed  of  a  plate  web  and  angle  flanges  which  are  riveted  to 
the  edges  of  the  plate  as  shown  at  (a),  Fig.  145.  Yet  in  addition  to  these 
parts  there  are  usually  vertical  angles,  known  as  stiffeners,  riveted  to  the 
web,  as  shown  at  (6),  to  stiffen  the  web  against  buckling.  The  stiffeners 
are  either  bent  around  the  flange  angles  as  shown  at  (c)  or  filler  plates 
are  placed  between  them  and  the  web,  as  shown  at  (d).  When  they  are 
bent  around  the  flange  angles  we  say  they  are  crimped.  It  is  practice  to 
limit  the  thickness  of  metal  to  about  f  of  an  inch  and  when  the  area 
required  in  a  flange  is  greater  than  that  of  two  6"  x  6"  x  f "  Z  s,  plates  are 
riveted  to  the  backs  of  the  outstanding  legs  of  the  flange  angles  as  shown 
at  (e),  to  provide  for  the  additional  area  required.  These  plates  are 
known  as  cover  plates,  or  flange  plates.  The  area  of  the  cover  plates  in 
any  flange  should  never  be  much  greater  than  that  of  the  two  flange 
angles,  and  when  they  exceed  the  area  of  the  angles  very  much,  plates, 
known  as  side  plates,  are  placed  between  the  flange  angles  and  the  web 


178 


STRUCTURAL  ENGINEERING 


as  shown  at  (/)  so  that  the  area  of  the  cover  plates  can  be  reduced.  IE 
this  case  the  side  plates  are  considered  as  part  of  the  flange.  There  are 
other  types  of  flanges  used  occasionally  which  will  be  shown  later. 


Web 


(a) 

•Shffeners 


o  0)0  ooo 

CJ  «/ 
II 

3H^O-e\o  -<j 

U.    ^ 

OOQO  O   <) 
o 

Web      {] 

o 

h  n  n  n  on 

O 

(c)      (d) 


(b) 

Cover  Plafes 


Fig.   145 

112.     Stress  and  Area  in  Flange. — In  case  of  a  beam  composed  of 
one   piece,  as   an   I-beam,   the   stress   due  to   cross   bending  is   obtained 
through    the    application    of    For- 
mula D,  Art.   53,  but  in  the  case  r 
of  a  plate  girder,  while  the  same 
method   would   apply,  quite  a   dif- 
ferent one  is  used,  wherein  the  web 
and  flanges  are  treated  separately. 
The   cross   bending  is   resisted   by 
the  flanges  and  web  combined,  but 
the  resistance  of  the  web  is  ignored 
by    some    engineers,    while    others 
consider  it.     So  we  really  have  two 
cases  to  consider. 

In  case  the  resistance  of  the 
web  be  ignored,  the  stresses  in  the 
two  flanges  form  a  couple  which 
will  be  equal  and  opposite  to  the 

algebraic  sum  of  the  external  cou-  Fig.  i46  " 

pies  on  either  side  of  any  cross-section — which  is  the  same  thing  as  the 
bending  moment. 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDERS  179 

Let  Fig.  146  represent  a  portion  of  a  plate  girder: 
Let  M  -  bending  moment  at  section  cc; 

F  =  total  stress  in  each  flange  at  that  section  ; 
d  =  distance  between   the   centers   of   gravity   of  the   flanges, 
which  is  known  as  the  effective  depth. 

Then,  in  accordance  with  the  condition  of  equilibrium,  we  have 

M  =  Fd  or  F  =  ^-  ....................................  (1). 

If  /  be  the  allowable  unit  stress,  the  area  required  for  each  flange  will  be 


This  is  assuming  that  the  unit  stress  is  the  same  over  the  entire  cross- 
section  of  each  flange  —  an  assumption  invariably  made,  although  not 
absolutely  true  in  any  case,  but  quite  accurate  enough  for  practical 
designing. 

In  case  the  resistance  of  the  web  be  considered,  the  bending  moment 
will  be  resisted  by  the  flanges,  the  same  as  shown  above,  aided  by  the 
web,  which  is  really  a  rectangular  beam.  Then  for  the  bending  moment 
we  have 


2 

where  /'  represents  the  unit  stress  on  the  extreme  elements  of  the  web, 
and  h  and  /  represent,  respectively,  the  height  and  moment  of  inertia  of 
the  web,  while  the  other  letters  signify  the  same  as  they  do  in  equation 
(2),  except  F,  of  course,  is  less. 

Now,  let  t  be  the  thickness  of  the  web,  A'  its  area  of  cross-section 
and  A  the  area  of  each  flange;  then  we  have 

A'=th,I  =  ±-ths 

IxJ 

and  as  the  unit  stress  on  the  extreme  elements  of  the  web  is  practically 
the  same  as  that  of  the  flanges,  we  have  /'=/  and  also  F  —  fA  =fA.  Then 
by  substituting  these  values  in  (3),  we  have 


But  h  =  d,  practically,  so  we  have 

fA'd        ,tA      A'\ 
M  =  fAd  +  '—  =fd  lA  +  —  \ 

from  which  we  obtain 


which  shows  that  one-sixth  of  the  area  of  the  web  appears  as  flange  area, 
but  owing  to  the  moment  of  inertia  of  the  web  being  reduced  in  most  cases 


180  STRUCTURAL  ENGINEERING 

by  rivet  holes,  one-eighth  is  assumed  in  practice  instead  of  one-sixth,  in 
which  case  we  have 

A'     M 
A  +  ^  =  fd  .......................    ..............  —  «• 

from  which  we  obtain 


for  the  area  required  in  each  flange. 

113.  Economic  Depth.  —  It  is  seen  from  the  preceding  article  that 
the  area,  and  consequently  the  weight,  of  the  flanges  of  a  plate  girder 
varies  inversely  as  the  depth  of  the  girder  and  it  is  evident  that  the 
weight  of  the  web,  stiffeners,  and  fillers  varies  directly  as  the  depth.  Then 
evidently  a  plate  girder  will  have  a  theoretical  economic  depth  when  the 
weight  of  the  two  flanges  equals  the  combined  weight  of  the  web,  stif- 
feners,  and  fillers.  There  are  really  two  cases  to  consider,  which  are  as 
follows  : 

Case  I.     When  the  web  is  not  considered  to  resist  cross  bending. 
Let  M  =  maximum  bending  moment  in  inch  pounds  ; 
L  =  length  of  girder  in  feet; 
/  =  allowable  unit  stress  on  flanges  ; 
t  =  thickness  of  web  in  inches; 
W  =  total  weight  of  girder  in  pounds  ; 

x  —  depth  of  girder  in  inches  back  to  back  of  flange  angles. 
Girders  without  cover  plates.     For  the  area  of  the  cross-section  of 
the  two  flanges  we  have 

2M 

f* 

assuming  depth  and  effective  depth  as  being  equal,  and  for  the  weight  of 
the  two  flanges  we  have 

2fx3.4L. 
fx 

As  a  bar  of  steel  one  square  inch  in  cross-section  and  one  foot  long 
weighs  3.4  pounds,  the  total  weight  of  the  web  alone  is  equal  to  ZALtx 
and  the  total  weight  of  the  stiffeners,  fillers,  splices,  etc.,  usually  runs 
about  60  per  cent  of  the  weight  of  the  web,  so  the  total  weight  of  the  web, 
stiffeners,  fillers,  etc.,  can  be  taken  as  1.6(3.4Lto). 

Now  adding  the  weight  of  the  web,  stiffeners,  fillers,  etc.,  to  that  of 
the  flanges,  we  have 

W  =  2-x  3.4L  +  1.6(3.4Lte)    ..........................  (1) 


for  the  total  weight  of  the  girder. 
This  will  be  a  minimum  when 


=  ~2     x  3AL  +  1-6(3-4L*)=  o. 


which  is  obtained  by  differentiating  (1). 


DESIGN  OP  I-BEAMS  AND  PLATE  GIRDEKS 
From  this  we  obtain  for  the  economic  depth 

............  .........  (2). 

Girders  with  cover  plates.  In  this  case  the  area  of  the  flanges  varies 
from  the  center  to  the  ends  of  the  span.  If  the  cover  plates  have  theo- 
retical lengths,  the  average  cross-section  of  each  flange  will  be  about  0.75 
of  the  maximum  area.  So  for  the  total  weight  of  the  two  flanges  we  have 


Now  substituting  this  instead  of  2(M/fx)  x  (3.4L)  in  (1)  and  differ 
entiating  and  reducing  we  obtain  for  the  economic  depth 

O  /QN 


Case  II.      When  the  web  is  considered  to  resist  bending  moment. 

By  assuming  the  web  to  resist  bending  moment  each  flange  can  be 
reduced  in  area  to  the  amount  of  one-eighth  of  the  area  of  the  cross- 
section  of  the  web,  and  hence  the  total  weight  of  the  girder  would  be 
reduced  an  amount  equal  to  2(^tx)3AL. 

Then  subtracting  this  from  (1)  we  have 


W  =  2    j      3AL  +  1.6(3.4Lte)-       i*    3.4L  ..............  (4) 

for  the  total  weight  of  the  girder.      Differentiating  this   equation   and 
placing  the  first  derivative  =  0,  and  reducing,  we  obtain 


/KN 


for  the  economic  depth  of  girders  without  cover  plates,  and  by  substituting 
1.5(M//j?)  x  (3.4L)  for  2(M/fx)  x  (3.4L)  in  (4)  and  differentiating  and 
placing  the  first  derivative  =  0,  and  reducing,  we  obtain 


x  = 


(6) 


for  the  economic  depth  of  girders  with  cover  plates. 

The  theoretical  economic  depth  of  plate  girders  can  be  computed 
from  the  above  formulas.  An  inch  or  two  either  way  from  the  theoretical 
depth  will  affect  the  design  but  little. 

114.  Length  of  Cover  Plates.  —  A  flange  of  a  plate  girder  varies  in 
area  along  the  girder  practically  as  the  ordinates  of  a  parabola,  the  same 
as  the  bending  moment.  (See  Art.  56.)  Then,  if  AB  (Fig.  147)  repre- 
sents the  length  of  a  plate  girder  and  OC  the  total  area  of  the  cross- 
section  of  one  flange  at  the  center  of  the  span,  any  ordinate  x  to  the 
parabola  ABC  will  represent  (approximately)  the  area  of  the  flange  at 
that  point. 

Suppose  the  flange  of  this  girder  to  be  composed  of  two  angles  and 
three  cover  plates, 


182 


STRUCTURAL  ENGINEERING 


Let al  =  area  of  top  plate; 

a2  =  area  of  second  plate  from  the  top; 
03  =  area  of  third  plate  from  the  top; 
a4  =  area  of  the  two  angles ;  and 
A  =  total  area  of  the  flange. 

Theoretically,  net  areas  should  be  used  throughout  and  one-eighth  of 
the  area  of  the  web  should  be  included  with  the  area  of  the  angles,  but  to 
provide  against  discrepancies  that  may  result  by  assuming  that  the  bend- 
ing moment  varies  along  the  girder  as  the  ordinates  to  a  parabola  (which 
is  not  absolutely  true  in  the  case  of  concentrated  loads),  we  will  use  gross 
areas  throughout  and  neglect  the  one-eighth  of  the  web  in  determining 
the  length  of  cover  plates. 


2-6 


oz 


^L 


03 


4 


^ 


B 


Fig.   147 


Now,  if  these  areas  be  laid  off  to  scale,  on  the  line  OC  (Fig.  147), 
the  theoretical  diagram  of  the  cover  plates  can  be  drawn  as  shown  and 
then  the  theoretical  length  of  each  plate  can  be  determined  by  scale.  In 
this  manner  the  length  of  the  cover  plates  on  any  plate  girder  can  be 
graphically  determined,  but  in  practice  the  lengths  are  usually  computed 
from  the  formulas  given  below,  as  that  is  the  more  convenient  way. 

In  accordance  with  the  properties  of  the  parabola  we  have,  referring 
to  Fig.  147, 


A 


from  which  we  obtain 


for  the  half  length  of  the  top  cover  plate,  and  multiplying  this  by  2,  we 
have 


(i) 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDERS  183 

for  the  full  theoretical  length.     Likewise  we  have 

#"       al  +  a2 


(I) 


from  which  we  obtain 


for  the  half  length  of  the  second  cover  plate  from  the  top,  and  multiplying 
this  by  2,  we  have 


lal  +  a2 


for  the  full  theoretical  length  of  that  plate. 
In  the  same  manner  we  obtain 


al  +  a2  +  a3  ,o\ 

-J- 

for  the  full  length  of  the  third  cover  plate  from  the  top,  and  further,  we 
have 


al  +  a2  +  a3  +  a4      T 

~    ~        =L 


for  the  full  length  of  the  angles. 

Now  from  the  above  it  is  readily  seen  that  the  general  formula  for 
the  theoretical  length  of  cover  plates  can  be  written  as 


-  '+an      ............................  (a). 


A 

Then  for  the  length  of  the  first  or  outside  plate,  in  either  the  top  or  bottom 
flange,  we  have 


for  the  second  plate  from  the  top  or  bottom  we  have 


"_  r 


and  for  the  third  we  have 


and  so  on. 

The  theoretical  lengths  of  cover  plates  can  be  determined  very 
readily  by  the  use  of  the  ordinary  slide  rule.  When  a  slide  rule  is  used, 
Formula  (a)  should  be  squared,  thereby  obtaining 

.  .  .  +an). 


184 


STRUCTURAL  ENGINEERING 


Then,  first  the  quantity  L?/A  can  be  set  off  on  the  upper  scale,  once  for  all. 
By  multiplying  this  by  al,  (al  +  a%),  (al  +  a2  +  a3),  and  so  on,  we  obtain, 
respectively,  the  square  of  the  lengths  of  the  first,  second,  third,  etc., 
cover  plate,  and  at  the  same  time  the  square  root  of  this  in  each  case 
(which  is  the  length  desired)  is  read  off  on  the  bottom  scale.  Thus  the 
length  of  each  cover  plate  on  a  girder  can  be  determined  by  one  setting  of 
the  rule. 

115.    Increment  of  the  Flange  Stress. —  Imagine  the  web  of  the 
girder  shown  at  (a),  Fig.  145,  to  be  made  up  of  vertical  strips  each  of 


r 


a  d 


ot.-oi-o-r-o  i  o 


J     J     i 


o     o 


o    o  r 


O   i  O  JO» 


Kl«^e 


4  I E  i 


,  o»i  o; 


B 


Rl 


Fig.   148 


which  is  connected  to  each  flange  by  one  rivet,  as  shown  in  Fig.  148. 

Suppose  the  girder  supports  a  number  of  loads  and  let  R  and  721  be 
the  reactions  due  to  these  loads  and  for  the  sake  of  simplicity  suppose 
both  of  the  reactions  and  all  of  the  loads  to  be  applied  directly  to  the  web. 

The  loads  tend  to  move  the  girder  downward  as  a  whole ;  this  motion 
is  prevented  by  the  reactions,  and  through  the  balancing  of  this  activity 
results  first  a  vertical  shearing  couple  on  each  of  the  imaginary  strips 
which  tends  to  rotate  each  strip,  but  rotation  of  each  is  prevented  by  the 
rivets  connecting  it  to  the  flange  angles,  whereby  the  flange  stress  results. 
To  show  this,  let  us  first  consider  the  end  strip  at  A.  The  shearing  couple 
on  that  strip  tends  to  rotate  it  clock-wise  which  causes  the  strip  to  exert 
(through  the  connecting  rivet)  a  force  to  the  right  upon  the  top  flange 
angles  and  an  equal  force  to  the  left  upon  the  bottom  flange  angles,  and 
the  same  is  true  of  the  second  strip  from  the  end  A,  and  of  the  third,  and 
so  on,  while,  as  is  readily  seen,  the  shearing  couple  on  each  of  the  strips 
near  end  B  tends  to  rotate  each  of  those  strips  in  the  opposite  direction,  or 
counter  clock-wise,  and  hence  the  force  exerted  on  the  top  flange  angles 
by  each  will  be  to  the  left,  while  the  equal  force  in  each  case  will  be 
exerted  to  the  right  upon  the  bottom  flange  angles. 

The  end  rivet  in  the  top  flange  angles  at  A  and  the  corresponding  end 
rivet  at  B,  acting  toward  each  other,  produce  a  simple  compressive  stress 
in  the  flange  angles  throughout  the  distance  between  these  rivets,  and, 
similarly,  the  second  rivets  from  the  ends,  acting  toward  each  other,  will 
produce  a  like  stress  in  the  angles  throughout  the  distance  between  those 
rivets,  and  the  third  rivets  from  the  ends  will  produce  a  like  compressive 
stress  in  the  flange  angles  from  one  rivet  to  the  other,  and  so  on.  The 
same  is  true  of  the  bottom  flange  except  the  direct  stress  produced  there  is 
tension. 

Thus  it  is  seen  that  the  stress  in  the  flanges  of  a  simple  plate  girder 


DESIGN    OP   I-BEAMS    AND    PLATE    GIRDERS  185 

is  increased  by  each  flange  rivet  as  we  pass  from  either  end  up  to  the 
point  of  maximum  moment.  (See  Art.  60.)  This  increase  at  each  rivet 
is  the  increment  of  the  flange  stress,  which  is  often  referred  to  as  the 
"flange  increment."  The  summation  of  these  increments  between  any  two 
points  would  be  known  as  the  increment  of  the  flange  stress  between  the 
two  points. 

116.  Spacing  of  Rivets  in  the  Vertical  Legs  of  Flange  Angles, 
—  The  rivets  in  a  plate  girder  are  practically  always  the  same  size 
throughout  the  girder,  in  which  case  the  rivets  in  the  vertical  legs  of  the 
flange  angles  should  be  so  spaced  that  the  pressure  against  the  flange 
angles  would  be  the  same  for  each  rivet  throughout  the  girder,,  and  hence 
the  stress  on  each  rivet  would  be  the  same  throughout.  So  the  problem 
involved  is  to  determine  the  pitch  of  the  rivets  so  that  the  increment  of  the 
flange  stress  is  just  equal  to  the  allowable  stress  on  the  flange  rivet  at  all 
points  along  the  flange. 

Let  S  be  the  shear  on  any  strip  abed  (Fig.  148)  of  longitudinal 
length  p,  and  let  r  represent  the  increment  of  the  flange  stress  at  that 
strip.  Now  as  the  couple  rh  is  equal  to  the  equal  and  opposite  couple 
reacting  on  the  strip  and  balancing  the  shearing  couple  Sp,  we  have 

Sp  =  rh, 
from  which  we  obtain 

rh  n^ 

P=^  .............................................  (1)- 

From  this  the  length  of  any  strip  can  be  computed  for  any  desired  value 
of  r.  But  the  length  of  any  strip  can,  in  all  cases,  be  taken  as  the  pitch 
of  the  rivets  at  the  same  point  and  hence  if  r  be  taken  as  the  allowable 
stress  on  one  rivet,  the  required  pitch  of  the  rivets  in  the  vertical  legs  of 
the  flange  angles  at  any  point  can  be  computed  from  this  formula. 

It  is  readily  seen  (from  Fig.  148)  that  these  rivets  would  fail  either 
in  double  shear  or  in  bearing  on  the  web,  and  r,  in  any  case,  should  be 
taken  equal  to  whichever  is  the  least.  The  bearing  on  the  web  is  usually 
the  least. 

There  are  really  four  cases  to  consider: 

Case  I.  When  the  loads  are  applied  directly  to  the  web  and  the 
resistance  of  the  web  to  bending  is  neglected.  In  this  case  the  pitch  of 
the  rivets  at  any  point  of  the  flange  is  determined  from  the  above  equation, 


where  p  =  pitch  ; 

r-  allowable  stress  on  one  rivet  in  either  double  shear  or  bearing 

on  the  web,  whichever  is  the  least  ; 

h  =  vertical  distance  between  the  rivets  in  the  two  flanges  ; 
S  -  shear  on  the  girder  at  the  point  considered. 

Case  II.  When  the  loads  are  applied  directly  to  the  web  and  the 
resistance  of  the  web  to  bending  is  considered.  In  this  case  a  certain  part 
of  the  shear  couple  Sp  is  resisted  by  the  web. 

Let  S  =  shear  on  the  girder  at  any  point  ; 

A  =area  of  the  cross-section  of  one  flange  at  any  point; 


386 


STRUCTURAL  ENGINEERING 


Af  =  area  of  the  cross-section  of  the  web; 
f  =  stress  per  square  inch  transmitted  to  each  flange  by  each 

rivet ; 

r  —  pressure  exerted  upon  the  flange  by  each  rivet ; 
h  =  vertical  distance  between  the  rivets  in  the  two  flanges ; 
p  =  pitch  of  flange  rivets. 

It  is  shown  in  Art.  112  that  the  resistance  of  the  web  to  cross  bending 
can  be  accounted  for  by  considering  one-eighth  of  its  area  as  being  con- 
centrated in  each  flange.  So  the  portion  of  the  shearing  couple  Sp  resisted 
by  the  web  can  be  approximately  expressed  as  fA'h/%  and  the  remaining 
portion  of  the  couple  which  is  resisted  by  the  flanges  can  be  approximately 
expressed  as  fAh.  Then  adding  these  two  expressions,  we  have 


But, 


then  substituting  this  value  of  /  in  the  last  equation  and  reducing,  we  have 
rhi        A" 

-*V+a, 


P 


for  the  pitch  of  the  rivets. 

Case  III.  When  the  loads  are  applied  di- 
rectly to  the  flange  angles  and  the  resistance  of 
the  web  to  bending  is  neglected.  In  this  case  the 
loads  will  exert  a  vertical  force  upon  the  rivets  in 
addition  to  the  horizontal  force  considered  above. 
Let  v  (Fig.  149)  represent  the  vertical  force  per 
linear  inch  of  flange  and  r  the  horizontal  force  on 
the  rivet  (the  same  as  above).  Then  for  the 
resultant  force  we  have 


But  from  (2)  we  have 


Fig.   149 


•-* 


and  substituting  this  in  the  last  equation,  we  have 


Then  transposing  and  extracting  the  square  root,  we  have 

R 
P  = 


S 
h 

for  the  pitch  of  the  rivets,  where 

p  =  pitch ; 

S  =  shear  on  the  girder  at  point  considered; 

v  —  load  on  the  flange  angles  per  linear  inch ; 


(4) 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDERS 


187 


R  =  stress  on  rivet,  which  should  be  taken  as  the  allowable  stress  on 

one  rivet; 
h  =  vertical  distance  between  the  rivets  in  the  two  flanges. 

Case  IV .  When  the  loads  are  applied  directly  to  the  flange  angles 
and  the  resistance  of  the  web  to  bending  is  considered.  We  have  here, 
the  same  as  in  Case  III, 


except 


r  — 


Sp 


which  is  obtained  from  (3). 

Then  substituting  this  value  of  r,  we  have 


from  which  we  obtain 


(5) 


for  the  pitch,  where  the  letters  signify  the  same  as  specified  above. 

In  case  the  flange  angles  have  double  rows  of  rivets,  h  should  be 
taken  as  the  mean  vertical  distance  between  the  rivets  in  the  two  flanges. 

117.  Web  Splice. —  It  is  always  desirable  that  the  web  of  any  plate 
girder  be  one  continuous  piece,  but  the  length  of  such  plates  is  limited  by 
the  steel  mills  and  whenever  the  length  of  a  web  exceeds  these  limits  it 
becomes  necessary  to  splice  it. 


8 


Fig.   150 


In  practice  there  are  two  standard  ways  of  splicing  a  web,  both  of 
which  are  shown  in  Fig.  150.     The  splice  shown  at  (a)  is  made  up  of  six 


188  STRUCTURAL  ENGINEERING 

splice  plates,  three  on  each  side  of  the  web,  while  the  splice  shown  at  (6) 
is  made  up  of  only  two  plates,  one  on  each  side  of  the  web.  The  splice 
plates  should  have  sufficient  material  to  transmit  the  shear  at  the  point  of 
splice,  and  also  the  cross  bending  on  the  web.  The  maximum  shearing 
and  maximum  bending  stress  at  any  point  in  a  simple  girder  do  not  occur 
at  the  same  time,  as  is  readily  seen,  yet  in  designing  web  splices  it  is 
practice  to  consider  the  two  occurring  simultaneously  at  the  splice — which 
is  an  assumption  entirely  on  the  side  of  safety. 

In  the  case  of  the  splice  shown  at  (a)  the  plates  marked  pi  are 
usually  assumed  to  resist  only  the  cross  bending  on  the  web,  while  the 
plates  marked  p2  are  assumed  to  resist  only  the  shear  on  the  web.  That 
is,  the  plates  pi  and  their  connection  to  the  web  are  designed  as  if  no 
shear  occurred  at  the  splice  and  plates  p2  and  their  connection  to  the  web 
are  designed  as  if  no  cross  bending  occurred.  To  illustrate  this  method  of 
designing  the  splice: 

Let /  =  allowable  stress  per  square  inch  in  each  flange; 
A'  =  area  of  the  cross-section  of  the  web; 

d  =  vertical  distance  between  the  centers  of  gravity  of  the  flanges ; 
d'  —  vertical  distance  center  to  center  of  plates  pi ; 
F  =  stress  in  each  pair  of  plates  pi ; 
S  =  maximum  shear  at  the  splice. 

Then  the  resistance  of  the  web  to  bending  is  represented  by  the 
couple  (fA'/8)d  which  must  be  equal  to  Fd'  and  hence  we  have 


8 

for  the  direct  stress  in  each  pair  of  plates  pi,  due  to  the  cross  bending  on 
the  web.  Now,  evidently,  these  plates  should  be  designed  to  take  this 
stress  F  and  the  number  of  rivets  on  each  side  of  the  splice  connecting 
each  pair  of  these  plates  to  the  web  should  be  sufficient  to  transmit  the 
stress  F  in  either  double  shear  or  bearing  on  the  web,  whichever  requires 
the  greater  number  of  rivets.  However,  as  the  bending  stress  varies 
directly  as  the  distance  out  from  the  center  of  the  web,  the  intensities  used 
in  designing  the  plates  pi  and  also  the  stress  allowed  on  the  rivets  con- 
necting them  to  the  web  must  be  proportional  to  the  intensities  allowed  in 
the  flange.  For  example,  if  the  allowable  stress  in  the  flange  is  /  per 
square  inch,  the  intensity  to  allow  on  the  plates  pi  would  be  (/)  d' /d, 
and  if  r  is  the  stress  allowed  on  a  rivet,  say,  in  bearing  on  the  web  at  the 
flange,  the  same  size  rivet  through  the  plates  would  have  an  allowable 
bearing  on  the  web  of  (r)  df '/d. 

Then  for  the  net  area  of  cross-section  of  each  pair  of  plates  pi  we 
have 

..I,-' 


And  for  the  number  of  rivets  required  on  each  side  of  the  splice,  we  have 

F 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDEKS  189 

In  designing  the  plates  p2,  all  we  have  to  do  is  to  select  two  plates 
that  have  sufficient  net  area  along  the  vertical  section  to  transmit  the 
shear  and  wide  enough  to  admit  the  necessary  rivets  on  each  side  of  the 
splice.  However,,  as  a  matter  of  fact,  there  is,,  as  a  rule,  more  metal  in 
the  two  splice  plates  than  is  necessary  to  carry  the  shear,  as  each  plate  is 
usually  as  thick  as  the  web,  in  which  case  the  two  plates  would  have  twice 
the  net  cross-section  of  the  web.  The  number  of  rivets  connecting  the 
plates  p2  to  each  side  of  the  splice  should  be  sufficient  to  transmit  the 
shear.  If  r  be  the  allowable  stress  on  each  rivet  —  and  we  will  assume 
that  each  resists  the  same  amount  of  stress  —  we  have 


for  the  number  of  rivets  on  each  side  of  the  splice  in  plates  p2.  The 
value  of  r  in  this  case  is  the  full  allowable  intensity  on  a  rivet  in  double 
shear  or  bearing  on  the  web,  using,  of  course,  whichever  is  the  least. 

The  assumption  that  plates  pi  resist  only  the  cross  bending  on  the 
web  and  plates  p2  only  the  shear  is  quite  a  reasonable  assumption  as 
plates  p2  are  too  near  the  center  of  the  web  to  resist  so  very  much  of  the 
cross  bending  and  plates  pi  are  really  so  narrow  and  too  far  from  the 
center  of  the  web  to  transmit  so  very  much  of  the  shear.  As  a  matter  of 
fact,  however,  the  plates  p2  do  resist  some  of  the  cross  bending  and  plates 
pi  resist  some  of  the  shear. 

In  the  case  of  the  splice  shown  at  (6)  the  two  plates  should  have 
sufficient  net  area  of  cross-section  along  the  line  cc  or  cfcf  to  resist  the 
cross  bending  on  the  web  and  also  the  shear. 

Let  /'  =  stress  per  square  inch  on  the  outer  edges  of  the  splice  plates  ; 

/  =  the  allowable  stress  per  square  inch  in  the  flanges; 
Af  —  area  of  cross-section  of  the  web; 
h  —  height  of  each  splice  plate; 

7  =  net  moment  of  inertia  of  the  two  splice  plates  along  the 
section  cc  or  c'  '  cf  (deducting  the  moment  of  inertia  of  the 
rivet  holes).  Then  we  have 


for  the  stress  per  square  inch  on  the  outer  edges  of  the  splice  plates.  This 
stress  should  not  exceed  fh/d.  As  a  matter  of  fact  it  never  does,  as  the 
two  splice  plates,  as  a  rule,  have  at  least  twice  as  much  area  of  cross- 
section  as  the  web.  However,  the  stress  /'  should  be  considered  in  doubt- 
ful cases.  The  net  area  of  the  cross-section  of  the  two  splice  plates  along 
section  cc  or  c'c'  should  be  sufficient  to  transmit  the  shear.  That  is,  the 
shear  divided  by  the  net  area  should  not  exceed  the  allowable  unit  shear- 
ing stress  on  the  web,  say  10,000  Ibs.  per  square  inch,  as  specified  in  the 
A.  R.  E.  Ass'n  Specifications. 

In  case  the  web  be  spliced  as  shown  at  (6)  we  can  not  well  consider 
other  than  that  the  maximum  stress  on  each  rivet  is  due  to  the  cross 
bending  on  the  web  and  shear  combined.  The  cross  bending  produces  a 
horizontal  force  on  each  rivet  while  the  shear  produces  a  vertical  force 
and,  as  is  evident,  the  maximum  stress  on  each  rivet  will  be  the  resultant 
of  these  two  forces.  The  vertical  shearing  stress  will  be  practically  the 


190  STRUCTURAL  ENGINEERING 

same  for  each  rivet,  but  the  stress  due  to  cross  bending,  which  (as  stated 
above)  is  transmitted  to  each  horizontally,  will  vary  directly  as  their 
distance  out  from  the  center  of  the  web. 

Let  s  be  the  horizontal  stress  due  to  cross  bending  on  each  of  the 
rivets  farthest  out,  as  indicated  at  (6),  and  let  e  be  the  distance  of  these 
rivets  from  the  center  of  the  web.  Then  if  there  were  a  rivet  out  unit 
distance  from  the  center  of  the  web  the  stress  on  it  would  be  equal  to  s/e 
and  evidently  the  stress  on  any  rivet  out  2  distance  from  the  center  of  the 
web  would  be  (s/e)  z  and  the  moment  of  this  force  or  stress  about  the 
center  of  the  web  would  be 


s  \  i 
-z  )=  - 
e  J  \t 


-   r 


Now  it  is  evident  that  if  this  moment  be  determined  for  each  and 
every  rivet  on  one  side  of  the  splice,  and  these  be  added  together,  their 
sum  would  be  equal  to  the  couple  (fA'/8)d.  So  we  have 

A*  = 


From  this  equation  s  can  be  determined  and  if  v  be  the  vertical  shear  on 
each  rivet,  which  is  readily  computed,  we  have 


for  the  maximum  resultant  stress  on  each  outer  rivet  which  is  the  absolute 
maximum. 

As  an  example,  let  a,  b,  c,  d,  and  e  represent,  respectively,  the  dis- 
tance of  the  first,  second,  third,  fourth,  and  fifth  horizontal  row  of  rivets 
out  from  the  center  of  the  web,  shown  at  (6),,  Fig.  150,  then  we  have 

n/2sa2     2sb2     2sc2     2sd*  \      ifA'\. 

2 (-    -+-     -  +  -    -  +  -     -  +  2se)  =    '—  )d. 
\    e  e  e  e  /       \  8  / 

From  this  we  obtain 

fA'd 


*  =    — 
32 


/a2      b2      c2      d2         \ 

I  —  +  —  4-  —  +  —  +  e  ) 
\e        e        e        e         I 


Then  combining  this  with  the  vertical  shear,  as  shown  above,  the  maximum 
stress  on  the  rivets  can  be  obtained.  However,  s  should  not  exceed  the 
allowable  at  the  flange  multiplied  by  2e/d. 

The  maximum  stress  on  the  rivets  in  the  type  of  splice  shown  at  (a) 
can  be  determined  in  the  same  manner.  In  fact,  the  rivets  in  all  web 
splices  should  be  tested  in  this  manner. 

118.  The  Stiffening  Angles  on  a  plate  girder  really  have  two  func- 
tions: one  is  to  stiffen  the  web  against  buckling,  and  the  other  is  to 
transfer  loads  from  the  flanges  directly  to  the  web.  It  is  obvious  that  the 
closer  the  stiff eners  are  spaced  along  a  web  the  stronger  the  web  is 
against  buckling,  and  consequently  the  higher  the  web  can  be  stressed. 
So  the  design  of  the  web  is  influenced  by  the  spacing  of  the  stiffening 
angles. 

There  is  no  theoretical  way  of  determining  the  spacing  of  stiff  eners. 
A  practical  rule  is  to  space  them  so  that  their  distance  apart  along  the 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDEHS  191 

girder  is  equal  to  the  depth  of  the  girder,  but  in  no  case  farther  apart 
than  five  feet  or  a  little  over.  However,,  to  satisfy  modern  requirements, 
it  is  usually  necessary  to  space  them  closer  together  near  the  ends  of  the 
girders  than  this  practical  rule  calls  for  in  order  to  obtain  an  economic 
web.  The  allowable  spacing  of  the  stiffeners  for  a  given  stress  on  the 
web  is  specified  as 


,,)  ....................................  (1) 

in  the  Specifications  prepared  by  the  A.  R.  E.  Ass'n, 
where  d  =  distance  between  the  stiffeners  (as  shown  at  (6),  Fig.  145); 
£  —  thickness  of  web; 

6'  =  shearing  stress  per  square  inch  in  the  web,  which  is  obtained 
by  dividing  the  shear  at  the  section  considered  by  the  gross 
area  of  the  cross-section  of  the  web. 
Equation  (1)  is  really  an  empirical  formula. 

It  is  not  economic,  as  a  rule,  to  space  stiffeners  closer  together  than 
half  the  depth  of  the  girder.  When  a  spacing  less  than  that  at  the  ends 
of  a  girder  is  given  by  the  above  formula,  a  thicker  web  should  be  used. 
There  should  always  be  stiffeners  at  any  point  where  excessive 
concentrated  loads  are  applied.  These  stiffeners  should  be  designed  as  a 
column  to  support  the  load,  the  effective  length  being  taken  as  half  the 
depth  of  the  girder. 

119.  Example  1.  —  Let  it  be  required  to  design  a  girder  30  ft.  long 
(c.c.  end  bearings),  to  support  a  uniform  dead  load  of  400  Ibs.  per  linear 
foot  of  girder,  and  a  live  load  of  4,000  Ibs.  per  foot. 

For  the  maximum  bending  moment  we  have  from  dead  load 

M  =  -i  x  400  x  302  x  12  =  540,000"  Ibs. 

8 

and  from  live  load 

M'  =  -  x  4,000  x"302  x  12  -  5,400,000"  Ibs., 

8  . 

making  a  total  of  5,940,000"  Ibs.  bending  moment. 

Let  us  assume  the  web  to  be  f  "  thick,  and  that  the  loads  are  applied 
directly  to  the  top  flange,  and  that  the  web  resists  bending.  Having  made 
these  assumptions  we  can  proceed  with  the  determination  of  the  economic 
depth.  Formula  (5),  Art.  113,  will  be  used,  as  girders  of  such  short 
lengths  usually  have  no  cover  plates.  So,  substituting  in  this  formula  we 
have 


1,940,000        OOA  ing 


>,000  x  3/8 

for  the  economic  depth  of  the  girder.  So  we  will  assume  a  38"  x  f "  web. 
The  next  thing,  we  will  test  this  web  to  see  if  it  is  satisfactory.  For  the 
maximum  end  shear  we  have  from  dead  load 

£  =  400x15  =  6,000  Ibs. 


192  STRUCTURAL  ENGINEERING 

and  from  live  load 

£'  =  4,000x15  =  60,000  Ibs., 

making  a  total  of  66,000  Ibs.  end  shear. 

Now    dividing    this    by    the    area    of    cross-section    of    the    web 
(=38"xf"),wehave 


for  the  actual  average  unit  shearing  stress  on  the  web. 

Next,  substituting  this  in  Formula  (1),  Art.  117,  we  have 

d  =  ^r  (12,000  -4,630)=  69.2  ins. 

for  the  maximum  theoretical  distance  between  stiffeners  at  the  ends  of  the 
girder;  and  as  this  spacing  is  not  less  than  half  the  depth  of  the  girder 
the  web  itself  is  satisfactory,  but  common  practice  is  not  to  permit  the 
distance  between  the  stiffeners  to  be  greater  than  the  depth  of  the  girder. 
So  we  will  space  them  so  as  not  to  violate  this  practice,  that  is,  the  clear 
horizontal  distance  between  the  stiffeners  will  not  be  made  more  than  38" 
in  any  case. 

Then  substituting  38  for  d  in  Formula  (1),  Art.  117,  we  have 

38=1(12,000-.), 

from  which  we  obtain 

s  =  7,946  Ibs. 

as  the  allowable  unit  shearing  stress  on  the  web. 
Now  dividing  this  into  the  shear  we  have 

66,000 


for  the  required  area  (of  cross-section)  of  the  web,  which  is  5.95n" 
(=14.25-8.3)  less  than  the  area  of  the  assumed  web.  So  theoretically 
the  web  can  be  thinner  than  §",  but  we  will  use  the  assumed  web  as  the 
specifications  limit  the  thickness  to  f".  However,  in  the  cases  of  build- 
ings and  highway  bridges,  the  web  would  likely  be  reduced  in  thickness, 
but  in  no  case  should  it  be  thinner  than  \"  . 

As  an  example  in  such  cases,  let  us  assume  a  38"  x  J"  web  which  has 
an  area  of  cross-section  of  9.5n"  (=38"x  J").  Then  for  the  actual  unit 
shearing  stress  on  the  web  we  have 

»  =  6,953  lb, 
J.o 

Substituting  this  in  Formula  (1),  Art.  117,  we  have 
d-  i  (12,000  -6,952)=  31.5  ins. 

for  the  required  spacing  of  the  stiffeners  at  the  end  of  the  girders,  and  as 


DESIGN  OF  I-BEAMS  AND  PLATE  GIRDERS  193 

this  is  not  less  than  half  the  depth  of  the  girder  the  web  is  satisfactory 
and  hence  would  be  used. 

We  will  next  take  up  the  designing  of  the  flanges.  Let  us  first  try 
2—  Ls  6"  x  6"  x  \"  for  each  flange.  From  Table  6  we  find  that  the  dis- 
tance from  the  back  of  these  angles  to  their  center  of  gravity  is  1.68". 
So  if  the  girder  be  made  38]"  deep  back  to  back  of  flange  angles,  that  is, 
J"  greater  than  the  depth  of  the  web,  which  is  usual  practice,  we  have 
38.25-1.68x2  =  34.89"  for  the  effective  depth.  Now,  by  dividing  this 
into  the  maximum  bending  moment  we  have 


for  the  stress  in  each  flange,  and  by  dividing  this  by  16,000,  the  allowable 
unit  stress,  we  have 

170,000 

10.6sq.ms. 


for  the  required  net  flange  area. 
We  have 

2_[_s  6"  x  6"  x  i"  =  11.50  -  1      =  10.50n"  net 
\  of  web  =  (14.25  x£)=  1.78n"net 

12.28n"net 
This  is  too  large,  so  let  us  try 

2__  [_s   6"  x  6"  x  Ty  =  10.12  -  .87  =  9.25n"  net 
1!rofweb  =  1.78n"net 

11.03n"net 

This  flange  section  is  about  right,  being  only  about  0.4n"  more  than 
required,  but  the  specifications  require  that  the  thickness  of  these  flange 
angles  be  at  least  one-twelfth  of  the  width  of  the  outstanding  legs  which 
are  6".  So  these  TV  angles  are  too  thin  and  consequently  if  6"  x  6" 
angles  be  used  at  all  the  6"  x  6"  x  -J"  angles  would  have  to  be  used,  which 
gives  an  excess  of  metal. 

Let  us  try  2—  Ls  6"x4"x^"  (the  4"  legs  outstanding)  for  each 
flange. 

The  distance  from  the  back  of  the  4"  leg  to  the  center  of  gravity  of 
each  of  these  angles  is  (given  in  Table  4)  1.99",  say,  2".     Then  for   fc 
effective  depth  we  have  38.25-2x2  =  34.25". 

Now  dividing  this  into  the  maximum  bending  moment  we  have 

5,940,000 


34.25 

for  the  stress  in  each  flange,  and  dividing  this  by  16,000  we  have 
173,000 


=  173,000  Ibs. 
[ividing  this  fr 
=  10.8  sq.  ins. 


16,000 

for  the  required  net  flange  area. 
We  have 

g_Ls  6"x4"xi"  =  9.50  -  1  =  8.50n"net 
£ofweb=  1.78Q"net 

10.28n"net. 


194  STRUCTURAL  ENGINEERING 

This  flange  section  is  0.52n"  less  than  the  required,  so  let  us  try  the 

following: 

2— Ls    6"  x  4"  x  TV  =  10.62  -  1.12  =  9.50n"  net 
J- of  web-  1.78n"nct 

11.28n"net. 

This  section  is  0.4Sn"  more  than  required.  So  either  of  the  last  two 
sections  may  be  used.  We  will  use  the  latter  (6"x  4"  x  -f$'f  angles)  as 
that  is  on  the  side  of  safety. 

The  size  of  the  intermediate  stiffeners  is  governed  practically  by  the 
specifications  which  require  that  they  be  no  less  than  §"  thick  and  the 
width  of  their  outstanding  legs  be  no  less  than  -£$  of  the  depth  of  the 
girder  plus  2  inches.  So  in  this  case  we  will  use  3J"  x  3 J"  x  f "  angles 
for  stiffeners  throughout  as  this  is  the  minimum  size  allowed.  Thickness 
of  the  end  stiffeners  would  very  likely  be  iV'  or  V >  depending  upon 
conditions.  If  the  girder  be  supported  upon  masonry,  these  stiffeners 
would  be  designed  as  columns  to  take  the  end  shear  or  reaction,  one-half 
of  the  depth  of  the  girder  being  taken  as  the  length  of  the  column. 

If  the  girder  be  riveted  at  the  ends  to  other  girders  or  to  columns 
which  wholly  support  the  girder,  the  end  stiffeners  in  that  case  would  be 
at  least  -faff  thick  to  provide  for  the  facing  of  the  ends  of  the  girder 
which  is  usually  required  in  modern  practice. 

To  obtain  the  spacing  of  the  flange  rivets  at  the  ends  of  the  girder, 
we  have 

R  =  7,880  Ibs.  allowable  bearing  of  a  J"  rivet  on  the  f"  web; 

t>  =  333  Ibs.  =  4,000 H- 12; 

£  =  66,000  Ibs.; 

h  =  31.5  =  (38.25  -  6.75). 

Then  substituting  these  values  in  Formula  (4),  Art.  116,  (the  formula 
required  by  the  specifications),  we  obtain 

7,880 

*"""  '66,000 


.or  the  pitch  of  the  flange  rivets  at  the  ends  of  the  girder.  This  pitch 
should  be  used  for  a  distance  out  from  the  ends  of  the  girder  equal  to  the 
depth  of  the  girder  and  varied  from  there  on  toward  the  center  of  the 
span  to  suit  the  shear  at  the  different  points,  no  pitch,  however,  being 
greater  than  6"  in  accordance  with  the  specifications. 

From  the  above  calculations  the  necessary  information  for  making 
the  detail  drawing  of  the  girder  is  obtained,  and  when  this  drawing  is 
finished  the  design  of  the  girder  is  complete.  All  plate  girders  are 
designed  in  this  manner. 


CHAPTER  XI 

DESIGN   OF    SIMPLE    RAILROAD   BRIDGES 

120.  Types. —  Simple   railroad   bridges    can   be   divided   into   four 
general  types:  beam  bridges,  plate  girders,  viaducts,  and  truss  bridges. 
Plate  girders   and  truss   bridges  can  be   further  divided  into  deck   and 
through  bridges;  deck  when  the  track  is   supported  on  the  top  of  the 
structure;  and  through  when  the  track  is  between  the  main  girders  or 
trusses.     Beam  bridges  and  viaducts  are  always  deck  bridges. 

121.  The  Specifications  prepared  by  the  American  Railway  Engi- 
neering Association,  referred  to  in  the  preceding  chapter,  will  be  used 
throughout  this  work. 

122.  The  Live  Load  usually  specified  for  railroad  bridges  consists 
of  two  typical  consolidated  locomotives  and  tenders  coupled  in  tandem 
followed  by  a  uniform  train  load,  and  a  special  load  concentrated  on  two 
axles  which  is  used  in  case  of  very  short  spans.     The  most  common  load- 
ing of  this  type  is  that  known  as  "Cooper's  Loading,"  devised  by  Theodore 
Cooper,  M.  A.  Soc.  C.  E.    Cooper's  loadings  vary  in  weight  and  are  desig- 
nated, beginning  with  the  lightest,  as  £30,  £35,  £40,  £45,  £50,  £55,  and 
£60.     These  loadings  vary  by  a  certain  ratio  so  that  if  any  stress,  shear, 
bending  moment,  etc.,  due  to  any  one  of  the  loadings  be  known  the  intensi- 
ties of  the  same  due  to  any  one  other  of  the  loadings  can  be  obtained  by 
direct  proportion,  and  hence  any  tables  or  diagrams  giving  the  shears, 
moments,  etc.,  for  any  one  of  the  loadings  can  be  used  for  any  of  the  other 
loadings.     The  figures  following  the  letter  £  are  the  indices  of  the  ratio 
referred  to  above.     Thus,  for  example,  if  any  stress,  shear,  etc.,  due  to 
£40  be  known,  the  intensity  of  the  same  for  £50  will  be  50/40  of  that 
due  to  £40;  60/40  for  £60;  35/40  for  £35;  and  so  on.     Most  of  the 
railroads  in  this  country  have  adopted  some  one  of  Cooper's  loadings, 
either  exactly  or  slightly  modified.     The  trend  has  been  toward  the  using 


Fig.   151 


of  the  heavier  loadings.  At  present  £50,  represented  in  Fig.  151,  is  used 
quite  extensively.  The  spacing  of  the  wheels  is  the  same  for  all  of  the 
loadings. 

The  £40  loading  is  the  most  convenient  to  use  owing  to  the  loads 
being  in  even  thousands  of  pounds. 

Table  A*  gives  for  this  loading  moments  about  any  wheel  of  the 
wheels  to  the  right  or  to  the  left  of  it,  as  will  be  seen  upon  inspection. 

*  The  student  should  verify  the  results   given  In  this  table. 

195 


196  STRUCTURAL  ENGINEERING 

This  table  is  quite  convenient  in  determining  the  centers  of  gravity, 
shears,  and  moments,  as  will  be  shown  later. 

123.  "An  Equivalent  Uniform  Live  Load"  is  sometimes  used 
instead  of  the  wheel  loads  described  above.  In  the  case  of  beams  and  deck 
girders  this  will  give  exactly  the  same  results  as  the  wheel  loads,  while  in 
the  case  of  trusses  and  through  girders  the  results  obtained  are  only 
approximately  the  same  as  for  the  wheels. 

To  obtain  an  equivalent  uniform  live  load  for  the  bending  moment 
on  a  beam  or  deck  girder  of  length  L,  the  maximum  bending  moment  M 
due  to  the  wheel  loads  is  computed,  and  then  we  have 


from  which  we  obtain 

8M 

**# 

for  the  equivalent  uniform  load  per   foot  which  will  produce  the  same 
maximum  moment  on  the  beam  or  girder  as  the  wheels. 

To  obtain  an  equivalent  uniform  live  load  for  the  end  shear  or 
reaction  on  a  beam  or  deck  girder  of  length  L,  the  maximum  reaction  R 
due  to  the  wheel  loads  is  computed  and  then  we  have 


from  which  we  obtain 


for  the  equivalent  uniform  live  load  which  will  produce  the  same  maxi- 
mum end  shear  or  reaction  as  the  wheels. 

In  practice  the  equivalent  uniform  live  load  for  trusses  is  usually 
taken  either  as  the  uniform  load  that  will  produce  as  great  a  bending 
moment  at  the  quarter  point  of  the  span  as  the  wheel  loads  or  the  uniform 
load  that  will  produce  as  great  a  shear  in  the  end  panel  as  the  wheel 
loads. 

The  first  is  obtained  by  computing  the  maximum  bending  moment 
M'  at  the  quarter  point  due  to  the  wheel  loads,  the  span  being  considered 
as  a  simple  deck  beam,  and  then  we  have 

-=¥• 

from  which  we  obtain 

32M' 


for  the  equivalent  uniform  live  load  per  foot  where  L  is  the  length  of 
span  in  feet. 

The  second  equivalent  load  is  obtained  by  computing  the  maximum 
shear  S  in  the  end  panel  due  to  the  wheel  loads  and  then  we  have 


-'  (¥> 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  197 

from  which  we  obtain 

S 
P=L^d 

2 

for  the  equivalent  uniform  live  load  where  L  is  the  length  of  span  and  d 
the  panel  length  in  feet. 

The  first  loading  gives  stresses  too  low  for  web  members  and  chords 
near  the  end  and  too  high  for  the  chords  near  the  center  of  the  span, 
while  the  second  loading  gives  stresses  about  correct  in  all  web  members, 
but  considerably  too  high  in  the  chord  members,  except  the  chords  in  the 
end  panels.  To  correct  for  this  discrepancy  the  author  has  proposed  the 
reduction  of  the  stresses  in  the  chords  (except  the  chords  in  the  end 
panels)  by  the  per  cent  given  by  the  following  empirical  formula:* 


where  L  is  the  length  of  span  in  feet. 

For  example,  the  chord  stresses  in  a  300-ft.  span  would  be  reduced 
by  5.5  per  cent.  This  does  not  apply  to  chords  in  end  panels. 

124.  Dead  Load  consists  of  the  weight  of  the  metal  in  the  structure 
(except  the  metal  at  the  supports),  and  the  weight  of  the  track,  known 
as  the  deck.  The  approximate  weight  of  metal  per  foot  of  single-track 
bridges  designed  for  Cooper's  E50  loading  can  be  obtained  from  the 
following  formulas  wherein 

p  =  weight  of  metal  per  foot  of  span, 
L  =  length  of  span  in  feet,  c.  c.  end  bearing: 
For  beam  spans  without  lateral  bracing,  and  stringers, 

p  =  l2L  +  100 (1), 

For  deck  girder  spans  and  beam  spans  with  lateral  bracing, 

p  =  12L  +  150 (2), 

For  through  plate  girder  spans, 

p  =  13L  +  600    (3), 

For  truss  spans, 

p  =  VL  +  660 (4) . 

The  weight  obtained  from  the  above  formulas  is  for  the  metal  alone, 
and  to  this  must  be  added  the  weight  of  deck  which  in  the  case  of  ordinary 
wooden  decks  can  be  taken  at  400  Ibs.  per  foot  of  track.  The  weight 
obtained  from  the  above  formulas  is  usually  correct  enough  for  computing 
stresses,  as  10  per  cent  variation  is  permissible,  but  it  should  not  be  used 
in  making  estimates  of  cost. 

The  approximate  dead  weight  of  metal  in  bridges  designed  to  carry 
Cooper's  E6Q  loading  is  about  one-eighth  more  than  given  by  the  above 
formulas,  and  the  weight  of  those  designed  to  carry  the  £40  loading  is 
about  one-eighth  less  than  that  given  by  the  above  formulas. 

The  weight  of  metal  in  double-track  bridges  depends  upon  their 
construction.  Their  weight  as  for  metal  is  about  70  per  cent  heavier  than 
that  of  single-track  bridges  if  three  or  two  main  girders  or  trusses  are 
used,  but  about  100  per  cent  if  four  trusses  or  girders  are  used. 

*  See  Engineering  News,  September  13,  1906. 


198  STRUCTURAL  ENGINEERING 

In  case  of  concrete  or  metal  floors  the  floor  should  be  designed  and 
then  the  weight  of  it  computed,  and  then  the  weight  of  the  main  girders 
or  trusses  can  be  assumed  and  added  to  this  weight  and  in  this  way  the 
approximate  dead  load  can  be  determined,  which  should  be  within  10  per 
cent  of  the  actual  weight;  if  not,  the  dead  load  stresses  should  be  redeter- 
mined,  using  a  revised  dead  load. 

125.  Impact. — Rapidly  moving  trains  will  produce  greater  stresses 
in  bridges  than  the  same  load  when  simply  standing  on  a  structure,  as  we 
consider  it  when  computing  the  live-load  stress,  and  to  provide  for  this 
additional  stress  a  certain  amount  of  the  live-load  stress  in  each  member 
is  taken  as  the  impact  stress  which  is  added  to  the  corresponding  live-load 
^tress.   The  impact  stress  is  obtained  by  simply  multiplying  the  maximum 
live-load  stress  by  a  coefficient  obtained  from  an  empirical  formula.     The 
following  formula  is  the  one  most  used: 

_300_ 
"L  +  300 

Then  for  the  impact  stress  in  any  member  we  have  the  formula 

r._o/    300    \ 
\L  +  300/ 

as  specified  by  the  A.  R.  E.  Ass'n  in  their  Specifications  referred  to  above, 
where 

7  =  impact  stress, 

S  =  maximum  computed  live-load  stress, 

L  =  length  of  track  (in  feet)  loaded  when  the  maximum  live-load 
stress  occurs. 

126.  Wind  Loads. —  The  horizontal  pressure  exerted  on  bridges  by 
the  wind  is  known  as  the  "Wind  Load."    As  a  rule  this  load  can  be  safely 
taken  at  30  Ibs.  per  square  foot  of  the  horizontal  projection  of  both  the 
structure  and  the  live  load  carried.     But,  in  addition  to  this,  some  pro- 
vision should  be  made  for  lateral  vibration  due  to  the  live  load.     To  pro- 
vide for  this  and  the  wind  load  a  greater  pressure  than  30  Ibs.  probably 
should  be  used  in  some  cases. 

The  lateral  force,  which  includes  the  wind  pressure,  specified  in  the 
A.  R.  E.  Ass'n  Specifications,  will  be  used  in  this  book. 

BEAM  BRIDGES 

127.  Preliminary. —  Beam  bridges  are  used  only  for  short  spans, 
rarely  ever  exceeding  20  ft.  in  length.     There  are  usually  at  least  two 
I-beams  under  each  rail  connected  to  each  other  by  diaphragms,  similar  to 
that  shown  m  Figs.  158  and  159,  thus  forming  a  compound  girder.    These 
compound  girders  should  be  braced  to  each  other  by  diagonal  and  trans- 
verse bracing,  as  shown  in  Fig.  159,  whenever  the  span  exceeds  12  ft. 
When  the  span  length  exceeds  15  ft.,  four  panels  of  bracing  should  be 
used,  in   which  case  a  horizontal  diaphragm  is   needed  at  each  lateral 
connection.      In  case  there  be  more  than  two  beams  in  each  compound 
girder  the  beams  can  be  placed  somewhat  closer  together  than  in  the  case 
of  two  beams. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


199 


Complete  Design  of  a  10-ft.  Span 

128.  Data.- 

Length  =  10'  c.c.  end  bearings. 
Dead  load  =  220 +  400  =  620  Ibs.  per  foot  of  span. 

(From  (1),  Art.  123.) 

Live  load,,  Cooper's  E5Q  (shown  in  Fig.  151). 
Specifications  A.  R.  E.  Ass'n. 

129.  Calculations. — For  the  maximum  bending  moment  due  to  dead 


load  we  have 


=  i  x  -^f-  x  102  x  12  =  46,500  inch  Ibs. 

o  & 


From  mere  inspection  of  the  live-load  diagram,  in  Fig.  151,  it  is  seen 
that  the  maximum  moment  will  occur  either  when  one  62,500-lb.  axle  load 
(of  the  special  load)  is  on  the  span  or  when  two  of  the  50,000-lb.  axle 
loads  are  on.  So  we  have  two  cases  to  try. 

The  maximum  moment  due  to  the  62,500-lb.  axle,  one-half  of  which 
goes  to  each  girder,  will  occur  (according  to  Art.  88)  when  the  load  is  at 
the  center  of  the  span,  as  shown  in  Fig.  152.  Each  of  the  reactions  will 


2.5' 


• 

?!    J 


31 


PB 


,t 


/O' 


Fig.   153 

be  equal  to  one-half  of  the  load  on  each  girder,  or  15,625  Ibs.     Then  we 
have 

15,625x5x12  =  937,500  inch  Ibs. 

for  the  maximum  bending  moment  due  to  this  load. 

The  maximum  due  to  the  two  50,000-lb.  axles  will  occur  when  the 
loads  are  in  the  position  shown  in  Fig.  153  (according  to  Art.  88)  and  it 
will  occur  under  the  wheel  marked  3. 

Taking  moments  about  A  (Fig.  153)  we  have 

Rl  x  10  -  25,000  x  1.25  -  25,000  x  6.25  =  0, 
from  which  we  obtain  the  reaction 

Rl  =  18,750  Ibs. 

Then  for  the  bending  moment  at  wheel  3  (which  is  the  maximum) 
we  have 

M'  =  18,750x3.75x12  =  843,700  inch  Ibs., 

which  is  less  than  that  produced  by  the  one  wheel  of  the  special  load,  as 


200  STRUCTURAL  ENGINEERING 

is  seen  above,  and  hence  the  moment  due  to  the  special  load  will  be  used. 
Then  for  the  maximum  impnct  moment  we  have 


=  937,500  =907,000  inch  Ibs. 

1U 


Now  adding  the  maximum  dead-  and  live-load  moments  and  impact 
together  we  have 

46,500  +  937,500  +  907,000  =  1,891,000  inch  Ibs., 

which   is   the  total  moment  which  the  girder  under   each   rail   must  be 
designed  to  resist. 

Dividing  this  by  16,000  (see  Art.  105)  we  have 

1,891,000 

—  TTTTTTTT;  —  =llo.«  tor  the  section  modulus. 

From  Table  I  we  find  that  this  calls  for  2—  Is  15"  x  42  #  under 
each  rail. 

We  can  now  make  a  preliminary  estimate  of  the  dead  load.  The 
weight  of  the  four  Is  is  168  Ibs.  per  foot  of  span,  and  the  details  can  be 
neglected  as  the  channel  diaphragm  (see  Fig.  158)  at  the  center  of  the 
span  is  all  the  detail  there  is  to  consider.  Then  adding  this  168  Ibs.  to 
the  weight  of  the  deck  we  have  568  Ibs.,  which  is  52  Ibs.  less  than  the 
assumed  dead  load,  but  as  this  is  within  10  per  cent  of  the  assumed  load 
no  recalculations  are  necessary. 

For  the  end  shear  or  reaction  due  to  dead  load  we  have 


R  =  — — —  x       =  1  550  Ibs 

The  maximum  end  shear  or  reaction  due  to  the  live  load  will  occur 
when  the  two  62,500-lb.  axles  are  on  the  span  and  in  the  position  indi- 
cated in  Fig.  154. 

Taking  moments  about  the  support  B  we  have 

R:  x  10  -  31,250  x  10  -  31,250  x  3  =  0, 

from  which  we  obtain 

^  =  40,600  Ibs.,  /T) 

D  T 

which  is  the  maximum  live-load  end  shear  or  reaction.      T 

For  the  impact  we  have  Fig.  154 


Now  adding  the  above  reactions  and  impact  together  we  have 

1,550  +  40,600  +  39,300  =  81,450  Ibs. 
for  the  total  end  shear  or  reaction.     Dividing  this  by  600  we  have 

81,450 

—  -  -  =136  sq.  ins. 
600 

for  the  required  area  of  bearing  on  the  masonry  for  each  of  the  four 
supports.  This  completes  the  necessary  calculations,  and  next  the  general 
drawing,  as  shown  in  Fig.  158,  or  a  shop  drawing  can  be  made  for  the 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


201 


span.     The  details  shown  in  Fig.  158  should  be  studied  by  the  student 
until  thoroughly  understood. 

Complete  Design  of  a  15-Ft.  Span 
Data.— 


130. 


131. 


Length  =  15'  c.c.  end  bearings. 
Dead  load  =  330  +  400  =  730  Ibs.  per  ft.  of  span. 

(From  (2),  Art.  123.) 
Live  load,  Cooper's  £50. 
Specifications,  A.  R.  E.  Ass'n. 
Calculations.  —  For  the  maximum  bending  moment  due  to  dead 


load  we  have 


=  -  x  —  x  152  x  12  =  123,000  inch  Ibs. 

8        2 


,  I 

Oil                        <Nll 

p±ms; 

R]2.5'| 

s'           s' 

2.SJK' 

r 

IS' 

-J 

It  is  readily  seen,  from  the  diagram 
in  Fig.  151,  that  the  heaviest  loading  pos- 
sible for  this  span  is  three  50,000-lb.  axle 
loads,  and,  according  to  Art.  88,  these  will 
produce  the  maximum  moment  when 
placed  in  the  position  shown  in  Fig.  155, 
where  the  loads  are  indicated  for  one 
girder  only. 
Then  taking  moments  about  B  (Fig.  155)  we  have 

R  x  15  -  25,000  x  2.5  -  25,000  x  7.5  -  25,000  x  12.50  =  0, 
from  which  we  obtain 

#  =  37,500  Ibs. 
for  the  reaction  at  A.  Then  taking  moments  about  the  center  load  we  have 

M'  =  (37,500  x  7.5  -  25,000  x  5)  12  =  1,875,000  inch  Ibs., 
and  for  the  impact  we  have 

300 


7=1,875,000 


1Kn 
lo  +  oUU 


=1,785,000  inch  Ibs. 


Now  adding  the  dead  and  live  moments  and  impact  together  we  have 
3,783,000  inch  Ibs.  for  the  maximum  bending  moment.  Then  dividing 
by  16,000  we  have 

3,783,000      _ 
16,000 

for  the  section  modulus  which  calls  for  2  —  Is  20"  x  70  #  under  each  rail. 
This  beam  appears  a  little  heavy,  from  the  section  modulus,  but  the 
material  cut  out  of  the  web  along  the  vertical  row  of  rivets  at  the  center 
of  the  span  must  be  taken  into  account.  This  is  done  by  subtracting  the 
moment  of  inertia  of  the  material  cut  out  of  the  web  from  the  moment  of 


202  STRUCTURAL  ENGINEERING 

inertia  of  the  beam.     Then  substituting  this  net  moment  of  inertia  in  the 
formula 

f=My 

which  is  Formula  D,  Art.  53,  we  obtain  the  maximum  stress  on  the  beam, 
which  should  not  exceed  16,000  Ibs. 

The  area  of  cross-section  cut  out  of  the  web  at  each  rivet  hole, 
assuming  that  |"  rivets  are  used,  is  it  x  A  =  0.46  square  inches.  Then 
taking  moments  about  the  neutral  axis,  which  we  will  assume  is  at  the 
third  rivet  from  the  bottom  of  the  beam,  which  is  only  approximately  true 
(see  drawing,  Fig.  160),  we  have 


0.46x6.75   +0.46x2.5   =23.8 

for  the  moment  of  inertia  of  the  material  cut  out  above  the  neutral  axis,, 
and 

0.46x^5  +0.46  x62  =  19.4 

for  the  moment  of  inertia  of  the  material  cut  out  below  the  neutral  axis. 
Then  adding  these  together  we  have  43.2  for  the  total  moment  of  inertia 
of  the  material  cut  out.  Subtracting  this  from  the  moment  of  inertia  of 
one  beam  we  have 

1219.9-43.2  =  1176.7 

for  the  net  moment  of  inertia  of  one  beam. 

Now  substituting  in  the  above  formula  we  have 

1      3,783,000x10 
/-8X"      1176.7        =16>1001bs"> 

which  is  very  nearly  the  allowable  stress.  So  the  beam  has  practically 
the  correct  section. 

The  rivet  holes  in  the  beams  shown  in  Fig.  158  are  so  near  the 
neutral  axis  that  the  moment  of  inertia  is  affected  but  little,  and  hence 
the  material  cut  out  was  not  considered  in  designing  those  beams. 

We  can  now  make  a  preliminary  estimate  of  the  dead  weight. 

The  four  beams  will  weigh  280  Ibs.  per  foot  of  span  and  the  laterals 
and  details,  including  lateral  connections  and  diaphragms,  will  be  about 
30  Ibs.  per  foot  of  span,  so  the  total  effective  weight  of  metal  per  foot  is 
about  280  +  30  =  310  Ibs.  Adding  this  to  the  weight  of  the  deck  we  have 
710  Ibs.,  which  is  20  Ibs.  less  than  the  assumed  dead  weight,  but  this  is 
much  less  than  the  allowable  10  per  cent  deviation,  so  recalculation  is 
unnecessary. 

The  weight  of  details  can  be  correctly  assumed  only  by  experienced 
designers.  It  is  necessary  that  the  student  draw  out  the  details  to  some 
extent  in  order  to  get  the  weight  of  metal  to  any  reasonable  degree  of 
accuracy. 

For  the  maximum  end  shear  or  reaction  due  to  dead  load  we  have 

rvqn 

R  =  -i-x  7.5  =  2,740  Ibs. 
II 

The  maximum  live-load  end  shear  or  reaction  will  occur  at  A  when 
the  wheels  are  iiv  the  position  shown  in  Fig.  156. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


203 


Then  taking  moments  about  B  we  have 

R'  x  15  -  (25,000  x  15  +  25,000  x  10  +  25,000  x  5)  =  0, 
from  which  we  obtain 

fl'  =  50,000  Ibs. 
for  the  maximum  live-load  end  shear  or  reaction.     For  the  impact  we  have 


=  50,000 


/      300     \ 
\15  +  300/ 


47,600. 


Adding  these  reactions  and  impact  together  we  have 

2,740  +  50,000  +  47,600  =  100,340  Ibs. 

for  the  maximum  end  shear  or  reaction. 
Dividing  this  by  600  we  have 

100,340 


600 


=  167.2  sq.  ins. 


for  the  required  bearing  on  the  masonry,  for  each  of  the  four  supports 
According  to  the  A.  R.  E.  Ass'n  Specifications  the  lateral  force  01 
the  laterals  will  be 

200  +  5,000x0.10  =  700  Ibs.  per  foot  of  span. 

This  force  is  applied  to  the  laterals  only  at  t!*e  central  connection.  Let 
Fig.  157  represent  the  plan  of  the  laterals.  There  will  be  a  load  of; 
700  x  7.5  =  5,250  Ibs.  applied  at  the  central  connection  which  must  be 


Fig.   156 


Fig.   157 


transmitted  to  the  ends  by  the  laterals,  one-half  going  each  way.  So, 
evidently,  the  maximum  shear  in  each  panel  will  be  2,625  Ibs.,  which  is 
one-half  of  5,250  Ibs.  The  stress  in  each  lateral  will  then  be  equal  to 
2,625  xsectf.  Sec(9  is  equal  to  about  2.6.  Then  we  have  2,625x2.6  =  6,825 
Ibs.  for  the  stress  in  each  lateral.  If  this  stress  be  tension,  as  it  would  be 
in  the  case  shown  in  Fig.  157,  the  net  area  required  in  each  lateral  would 
be  equal  to 

6,825 


but  when  the  lateral  force  is  exerted  in  the  opposite  direction  the  laterals 
would  be  subjected  to  stress  of  the  same  intensity,  but  it  would  be  com- 
pression, in  which  case  the  area  required  would  be  equal  to 

6,825 
( 16,000  -70- 


204  STRUCTURAL  ENGINEERING 

So  the  laterals  must  be  designed  as  compression  members  as  well  as 
tension  members.  Now  as  the  stress  is  so  very  small  the  ratio  of  length 
to  radius  of  gyration  will  really  govern.  This  should  not  be  over  ISO. 
As  regards  construction,  a  single  angle  for  each  lateral  is  the  best  section. 
Then  if  we  use  an  angle  the  first  question  is  —  which  radius  of  gyration 
shall  we  use  in  designing  it?  The  laterals  are  fixed  at  their  ends  in  the 
horizontal  plane  so  that  if  the  radius  about  the  vertical  axis  be  used,  only 
one-half  of  the  length  of  each  lateral  should  be  taken  as  the  length.  In 
the  vertical  plane  the  laterals  are  only  partially  fixed  at  their  ends,  as  the 
connection  plates  resist  bending  upward  and  downward  only  to  a  limited 
extent  so  that  if  the  radius  about  the  horizontal  axis  be  taken,  the  full 
length  of  the  lateral  should  be  taken,  which,  to  be  sure,  is  on  the  side  of 
safety.  As  the  laterals  are  fixed  in  the  horizontal  plane  and  partially 
fixed  in  the  vertical  plane  they  cannot  fail  readily  about  the  45°  axis,  so 
it  would  not  be  correct  to  use  the  least  radius  of  gyration  of  an  angle. 
From  this  it  is  seen  that  the  radius  about  the  horizontal  axis  is  the 
one  to  use.  The  length  of  each  of  the  laterals  is  about  8  ft.  or  96  ins. 
Then  the  radius  of  gyration  about  the  horizontal  axis  must  not  be  less  than 


which  permits  the  use  of  a  3"  x  3"  angle,  but  the  specifications  limit  use 
to  3J"  x  3J"  x  f  "  angles,  so  this  size  will  be  used. 

In  the  designing  of  the  transverse  struts,  sense  of  fitness  must  govern 
to  some  extent  as  rigidity  is  the  thing  sought.  However,  the  stress  is 
readily  determined.  The  one  at  the  center  of  the  span  has  the  greatest 
load  which  is  5^250  Ibs.,  which  produces  a  compression  stress  in  the  strut 
of  that  amount,  as  is  readily  seen  from  Fig.  157.  A  9"  [  is  the  maximum 
size  that  will  give  us  a  convenient  three-rivet  connection.  This  size  also 
looks  about  correct  as  far  as  rigidity  is  concerned. 

The  length  of  these  struts  is  about  42  ins.  and  the  least  radius  of 
gyration  of  the  9"  [  is  0.64.  Then  we  have 


which  is  quite  low.  So  the  9"  [  appears  to  be  satisfactory,  and  we  will 
use  1  —  [  9"x20#  for  each  transverse  strut. 

This  completes  the  necessary  calculations  and  next  a  general  draw- 
ing as  shown  in  Fig.  159  can  be  made.  This  drawing  shows  the  design  only 
in  a  general  way  and  the  bridge  could  not  be  fabricated  from  it.  Such 
drawings  are  usually  made  in  railroad  offices  and  in  the  offices  of  con- 
sulting engineers.  A  bridge  company  would  make  a  shop  drawing  as 
shown  in  Fig.  160,  and,  in  addition,  the  necessary  shop  bills  from  which 
the  bridge  would  be  fabricated. 

The  following  shop  bills  are  about  what  a  bridge  company  would 
require  for  the  work.  However,  bridge  companies  use  printed  forms. 

Page  No.  1  of  the  shop  bills,  which  would  accompany  the  shop 
drawing  shown  in  Fig.  160,  is  principally  for  the  templet  and  bridge  shop. 

Pages  No.  2  and  No.  5  are  principally  for  the  forge  shop. 

Page  No.  3  is  principally  for  the  pattern  shop  and  foundry. 

Pages  No.  4  and  No.  5  would  be  used  by  the  shipping  department. 


205 


Y////////////// 


206 


207 


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214 


STRUCTURAL  ENGINEERING 


The  superintendent  of  the  shops,  the  inspectors,  the  erectors  and 
others  would  receive  the  shop  drawing  and  bills,  depending  upon  the  way 
the  work  is  handled.  The  different  companies  do  not  have  exactly  the 
same  kind  of  bills  nor  do  they  handle  the  work  exactly  in  the  same  way. 
The  above  bills  are  intended  only  as  a  fair  example  of  shop  bills. 


DRAWING  ROOM  EXERCISE  NO.  2 

Design  a  12-ft.  I-beam  span  and  make  a  general  drawing  to  a  1" 
scale  and  a  tracing  of  the  same.  The  details  to  be  similar  to  those  shown 
in  Fig.  159. 


Data: 


Length  =  12'  0"  c.c.  end  bearings. 

Width- 5' 0"  c.c.  of  girders. 

Dead  load  to  be  assumed. 

Live  load,  Cooper's  E5Q. 

Specifications,  A.  R.  E.  Ass'n. 


Fig.   161 


A  layout  to  a  1J"  or  3"  scale  similar  to  the  one  shown  in  Fig.  161 
should  be  made  before  beginning  the  final  drawing  for  the  span  in  order 
to  determine  the  correct  dimensions  of  details. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


215 


DECK  PLATE  GIRDER  BRIDGES 

132.  Preliminary. —  Deck  plate  girder  bridges  are  used  for  spans 
ranging  from  25  ft.  to  110  ft.  There  have  been  a  few  spans  longer  than 
110  ft.  built,  but  such  lengths  are  not  economic,  as  a  rule. 

A  single-track  deck  plate  girder  bridge  is  composed  of  two  main 
girders  connected  to  each  other  by  cross  frames  and  laterals.  An  isometric 
view  of  a  typical  single-track  span  is  shown  in  Fig.  162  where  the  names 
of  the  different  parts  of  the  structure  are  given. 

Double-track  deck  plate  girder  spans  are,  as  a  rule,  composed  of  two 
single-track  spans  placed  side  by  side. 


Fig.  162 


Complete  Design  of  50-ft.  Single-Track  Deck  Plate  Girder  Span 
133.    Data.— 

Length  =  50'-0"  c.c.  end  bearings. 
Width  =  6'-6"  c.c.  girders. 

Dead  load  =  (750  +  400)  =  1,150  Ibs.  per  ft.  of  span  (Art.  124). 

Live  load,  Cooper's  ESQ. 
Specifications,  A.  R.  E.  Ass'n. 


216  STRUCTURAL  ENGINEERING 

134.  Calculations  for  Main  Girders. — For  the  maximum  bending 
moment  due  to  dead  load  we  have 

M  =  i  x  575  x~502x  12  =  2,156,000  inch  Ibs. 

o 

From  inspection  of  the  diagram  of  the  loading  in  Table  A,  we  can  see 
that  the  maximum  moment  due  to  live  load  will,  very  likely,  occur  under 
one  of  the  heavy  wheels,  either  wheel  12  or  13.  Let  us  assume  wheel  13, 
and  let  us  consider  wheels  9  to  16  on  the  span.  Taking  moments  about 
wheel  16  (using  Table  A)  we  have 


for  the  distance  that  the  center  of  gravity  of  these  wheels  (9  to  16 
inclusive)  is  to  the  left  of  wheel  16.  This  shows  that  the  center  of  gravity 
comes  2.24  ft.  to  the  left  of  wheel  13.  Then  the  maximum  bending 
moment  under  wheel  13  will  occur  when  the  loads  are  in  the  position 
shown  in  Fig.  163.  (See  Art.  88.) 

2-14     ,  ' 

Q  12  i  i?s$\*\-+i  i?  4.88 


8'  ©    8'  ©5®>!®5© 


25' 


Rl 


Fig.   163 


Taking  moment  about  B   (Fig.  163),  and  using  Table  A,  we  have 
R  x  50  -  2,740  -  (155  -  26)  4.88  =  0, 

from  which  we  obtain 

#=67,390  Ibs. 

for  the  reaction  at  A. 

Then  taking  moments  about  wheel  13,  considering  the  forces  to  the 
left,  we  have 

M'  =  67,390x26.12  -818,000  =  942,200  foot  Ibs. 

or  11,306,000  inch  Ibs.  for  the  bending  moment  at  that  wheel  when  wheels 
9  to  16  are  on  the  span.  Next,  suppose  wheels  10  to  16  on  the  span. 
Taking  moments  of  these  wheels  about  16  (using  Table  A)  we  have 

-       /  2,155  ' 

#-  = 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


217 


for  the  distance  that  the  center  of  gravity  of  the  wheels  (10  to  16 
inclusive)  is  to  the  left  of  wheel  16.  Therefore,  the  center  of  gravity  is 
0.43  ft.  to  the  right  of  wheel  13.  But  when  these  wheels  are  placed  on 
the  span  for  maximum  moment,  wheel  17  comes  on  from  the  right.  So 
we  will  consider  wheels  10  to  17  on  the  span.  Taking  moments  about 
wheel  17  we  find  that  the  center  of  gravity  of  wheels  10  to  17  is  2.9  ft. 
to  the  right  of  wheel  13.  So  this  group  of  wheels  will  be  in  the  position 
shown  in  Fig.  164  when  the  maximum  under  wheel  13  occurs. 


Fig.   164 

Then  taking  moments  about  B  (Fig.  164)   (using  Table  A)  we  have 

fix  50  -2,851  -(142  -13)  1.45-0, 
from  which  we  obtain 

fi  =  60,760  Ibs.  for  the  reaction  at  A. 
Then  taking  moments  about  wheel  13  we  have 

M"=  60,760x23.55  -480,000  =  950,900  foot  Ibs. 

or  11,411,000  inch  pounds  for  the  maximum  bending  moment  under  wheel 
13  for  this  group  of  wheels.  This  is  the  absolute  maximum  as  will  be 
found  by  further  trials.  It  is  seen  that  the  center  of  gravity  is  a  little 
nearer  to  wheel  14  than  it  is  to  13,  yet  the  maximum  moment  occurs  under 
wheel  13.  This  is  one  of  the  few  cases  where  the  maximum  does  not 
occur  under  the  wheel  nearer  the  center  of  gravity.  (See  Art.  88.) 
Generally,  the  position  of  the  wheels  for  maximum  moment  can  be  ascer- 
tained the  first  trial.  The  above  span  is  about  as  troublesome  as  any 
found. 

For  the  maximum  impact  we  have 


/=  11,411,000  x 


300 


L 

\50  +  300 


780,000  inch  Ibs. 


Now  multiplying  the  above  maximum  moment  and  impact  each  by 
50/40,  as  the  loading  specified  is  £50,  and  adding  these  results  to  the 
dead-load  moment,  we  have 

2,156,000+14,263,000  +  12,220,000  =  28,639,000  inch  Ibs. 
for  the  total  maximum  moment  on  the  span. 

The  next  thing  is  to  determine  the  economic  depth.  Assuming  the 
thickness  of  the  web  to  be  f"  and  substituting  in  equation  (6),  Art.  113, 
we  have 


=  1.055 


/28,< 

Vie: 


639,000 


000  x 


=  72.7  ins., 


so  we  will  use  a  72"  x  f  "  web. 


218  STRUCTURAL  ENGINEERING 

We  can  now  determine  the  flange  area  required.  It  is  readily  seen 
that  the  effective  depth  will  likely  be  a  little  less  than  the  depth  of  the 
web,  so  let  us  assume  71"  as  the  effective  depth.  Then  we  have 

38,639,000  +  71  =  403,000  Ibs. 
for  the  flange  stress. 
Then  403,000  + 16,000  =  25.2  sq.  ins., 

the  net  area  required  for  each  flange. 
Using  the  following: 

2— Ls  6"x6"xTy  =  12.86  -  2.25=  10.61n"  net 
1— cover  plate  14"  x  J"  =  7.00  -  1  =  6.00n"  net 
1— cover  plate  14"xTV'=  6.12  -  0.87  -  5.25n"  net 
I  of  the  web  =  3.37n"  net 

25.23°"  net 

for  each  flange  we  have  practically  the  required  area. 

We  can  now  determine  the  actual  effective  depth.  Let  Fig.  165 
represent  the  cross-section  of  the  above  flange. 


_  „ _       «r^ 

^p  .11.  /M.  a  *. 

NjJ 


oiar"  /?* 

21$  6*6  *j 


Fig.   165 

Then  taking  moments  about  the  center  of  the  top  cover  plate  (see 
Art.  47),  we  have 

-_  12.86x2.42  +  7.00x0.46  _ 

for  the  distance  from  the  center  of  the  top  cover  plate  to  the  center  of 
gravity  of  the  flange. 
Then  we  have 

1.32-0.71  =  0.61  ins. 

as  the  distance  from  the  back  of  the  angles  to  the  center  of  gravity  of  the 
flange. 

The  web  is  assumed  72"  deep  and  according  to  practice  the  vertical 
distance  from  the  back  of  top  flange  angles  to  the  back  of  the  bottom 
flange  angles  will  be  0.25"  more  or  72.25".  Then  for  the  actual  effective 
depth  we  have 

d=  72.25  -  (0.61  x  2)  =  71.03  ins., 

which  is  almost  what  we  assumed,  so  recalculation  of  the  flange  is 
unnecessary.  As  a  rule,  the  assumed  effective  depth  does  not  come  so 
close  to  the  actual ;  however,  \"  either  way  will  not  materially  affect  final 
results.  If  a  greater  difference  than  -J"  is  obtained  the  stress  and  section 
of  the  flange  should  be  recalculated,  using  the  computed  effective  depth. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  219 

The  thickest  cover  plates  should  always  be  placed  next  to  the  flange 
angles.  So  in  this  case  the  \"  plate  will  be  next  to  the  angles,  as  shown 
in  Fig.  165. 

For  the  length  of  the  -}$'  cover  plate,  or  outside  plate,  we  have 

1  50^5=24.3  ft.         ;;'  '      ' 

(see  Art.  114)  and  for  the  length  of  the  \"  cover  plate  we  have 


It  is  customary  to  make  cover  plates  from  two  to  three  feet  longer 
than  the  theoretical  length,  so  we  will  make  the  above  plates  27  and  37 
feet  long,  instead  of  24.3  and  35.5. 

The  next  thing  to  determine  is  the  stiffening  angles  and  the  web.  To 
do  this  the  first  thing  is  to  calculate  the  maximum  end  shear. 

For  maximum  reaction  or  end  shear  due  to  dead  load  we  have 


The  maximum  live-load  reaction  or  end  shear  will  occur  when  the 
wheels  are  in  the  position  shown  in  Fig.  166.     (See  Art.  86.) 

Z' 


Fig.   166 

Taking  moments  about  B,  Fig.   166,  and  using  Table  A,  we  have 

R'x50-4,072-142x2  =  0, 
from  which  we  obtain 

R'  =  87,100  Ibs. 

for  the  maximum  reaction  or  end  shear  at  A  due  to  JE40  loading.     Then 
for  £50  we  have 

87,100x^|  =108,900  Ibs. 
For  impact  we  have 


I  =  108,900  x          =  93,400  Ibs. 
o50 

Now  adding  the  above  dead-  and  live-load  reactions  and  impact 
together  we  have 

14,400  + 108,900  +  93,400  =  216,700  Ibs. 

for  the  maximum  reaction  or  end  shear  on  each  girder. 

According  to  the  specifications  the  outstanding  legs  of  intermediate 
stiffeners  must  not  be  less  than  one-thirtieth  of  the  depth  of  the  girder 
plus  two  inches.  So  we  have 

Z+ 2  =  4.4  ins. 


220  STRUCTURAL  ENGINEERING 

for  the  required  width  of  their  outstanding  legs.  The  standard  angle 
coming  nearest  to  this  requirement  is  a  5"  x  3J" ', 
which  will  be  used  throughout. 

^ie  enc^  stiff61161*8  must  be  designed  to  transmit 
the  total  maximum  reaction,  each  pair  being  consid- 
ered as  a  column  having  a  length  equal  to  one-half 
the  depth  of  the  girder.  (See  specifications.)  Then 
assuming  Ls  —  5"  x  3J"  x  \"  used,  we  have  for  each 
pair  a  column  having  a  cross-section  as  shown  in 
Fig.  167.  For  the  radius  of  gyration  of  this  column 
in  reference  to  axis  y-y,  we  have 


r=    /8.  x2.412  +  19.98 

Then  substituting  this  radius  and  36"  for  the  length  in  Formula  Q, 
Art.  73,  we  have 

16,000-70^  =  15,100  Ibs. 

for  the  allowable  compressive  unit  stress  on  the  end  stiffeners. 
Now  dividing  this  stress  into  the  maximum  reaction  we  have 

216,700 

=  14.00  sq.  ins. 


for  the  required  area  of  cross-section  of  the  end  stiffeners.     So  we  will 
use  two  pairs,  or 

4—  Ls  5"x3i"xl"  =  16.0  sq.  ins., 

the  5  x  3  J  x  -f^  angles  being  a  little  too  small. 

There  is  no  theoretical  way  of  computing  the  area  of  the  inter- 
mediate stiffeners.  It  is  practice  to  make  them  as  thin  as  the  specifica- 
tions will  permit.  So  we  will  use  5"  x  3J"  x  f  "  angles  throughout  for 
intermediate  stiffeners. 

Using  the  assumed  web,  72"  x  f  ",  we  have 

216,700 
s=  -  —  —  =8,020  Ibs. 

for  the  maximum  average  unit  shearing  stress  in  the  web. 
Then  substituting  in  Formula  (1),  Art.  118,  we  have 


40        '  >          =  37  ins'  (about) 

for  the   required  distance  between  the   stiffeners  near  the  ends   of   the 
girders.     Now  as  this  spacing  is  not  less  than  half  the  depth  of  the  web 
the  assumed  web  is  economic  and  hence  will  be  used. 
For  the  bearing  on  the  masonry  we  have 

216,700 

sq.ms. 


Each  support  must  be  so  designed  that  there  will  be  at  least  this  much 
area  of  bearing  on  the  masonry. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


221 


This  completes  the  necessary  calculations  for  the  main  girders,  as 
far  as  the  general  design  is  concerned,  and  the  next  thing  is  the  designing 
of  the  lateral  bracing,  that  is,  the  laterals  and  frames. 

135.  Calculations  for  Lateral  Bracing. — The  lateral  bracing 
should  always  be  symmetrical  about  the  center  of  the  span.  The  laterals 
should  have  a  slope  as  near  45°  as  is  practicable.  The  distance  between 
cross-frames  should  never  be  over  15  ft.  In  accordance  with  this  there 
must  be  three  intermediate  cross-frames  in  a  50-ft.  span,  as  a  less  number 
would  place  them  more  than  15  ft.  apart.  So  the  lateral  bracing  will  be 
as  shown  in  Fig.  168. 

so' 


®[ 


&         © 


r 

4-          3 

@    © 

Fig.   168 


r 

/ 

0 


According  to  the  specifications  the  laterals  must  resist  a  uniform 
lateral  force  of  200  +  0.10  (5,000)  =  700  Ibs.  per  ft.  of  span,  considered 
as  a  moving  load. 

Suppose  this  force  or  load  acts  from  the  direction  indicated  by  the 
arrows,  and  suppose  it  moves  on  to  the  span  from  the  right,  as  a  uniform 
live  load.  The  panel  load  at  any  point  h,  g,  etc.,  will  be 

P  =  700x6.2  =  4,340  Ibs.  (about). 

Then  for  the  maximum  shear  in  the  different  panels,  according  to 
Art.  90,  we  have 

Shear  in  panel  hg  =  ——-  x    1=       540  Ibs. ; 


x  3=  1,630  Ibs.; 
x  6=  3,260  Ibs.; 
xlO=  5,420  Ibs.; 
x!5=  8,140  Ibs.; 
x  21  =  11,400  Ibs.; 


«       «       «< 


28  =  15,200  Ibs. 


Now  if  each  of  these  shears  be  multiplied  by  the  secant  of  the  angle 
marked  Q  we  shall  obtain  the  stress  in  the  corresponding  diagonals  (see 
Art.  92). 


222  STRUCTURAL  ENGINEERING 

The  tangent  of  angle  0,  for  the  purpose  of  determining  stresses, 
can  be  taken  as 

||  =  0.962  (about), 

and  the  corresponding  secant  can  then  be  found  in  almost  any  table  of 
natural  trigonometrical  functions   (see  Carnegie  or  Cambria  handbook)  ., 
or  the  secant  can  be  determined  directly  by  arithmetic. 
For  the  above  assumed  figures  we  have 

Sec  6  =  1.39. 

Then  for  the  maximum  stress  in  the  diagonals  we  have 

5,420x1.39=-  7,500  Ibs.   for  diagonal  nd, 

8,140x1.39  =  +11,300  Ibs.  for  diagonal  dm, 

11,400x1.39  =  -15,800  Ibs.  for  diagonal  mb, 

15,200x1.39  =+21,000  Ibs.  for  diagonal  bl. 

These  are  all  of  the  stresses  that  we  need  determine  with  the  load 
applied  as  indicated  as  the  corresponding  diagonals  in  the  right  half  of 
the  span  will  have  the  same  stress  when  the  force  moves  on  to  the  span 
from  left  to  right. 

If  the  lateral  force  comes  from  the  other  direction  than  that  indi- 
cated by  the  arrows,  the  panel  loads  will  be  twice  as  great  as  given  above, 
as  there  are  only  half  as  many  panels  considered. 

Then  the  stress  in  each  of  the  diagonals  nd  and  dm  is 

4'34°  X  2  x  3  x  1.39  =  ±  9,050  Ibs. 


4- 
and.  -~     x  6  x  1.39  =  ±  18,100  Ibs. 

in  each  of  the  diagonals  mb  and  bl. 

We  now  have  the  maximum  stresses  in  the  diagonals  determined 
which  are  indicated  in  the  diagram  in  Fig.  168. 

The  plus  and  minus  signs  above  signify  compression  and  tension, 
respectively. 

It  is  seen  from  the  above  that  the  diagonals  have  to  resist  both 
tension  and  compression.  Compression  will  likely  govern. 

Let  us  try  a  single  angle,  say,  1  —  L  3J"  x  3  £"  x  f",  as  this  is  the 
smallest  that  the  specifications  will  permit.  From  Table  6,  or  from  a 
Carnegie  or  Cambria  handbook,  we  have  1.07  for  the  radius  of  gyration 
of  this  angle  about  the  horizontal  axis  (see  Art.  130)  and  the  length  of 
each  lateral  will  be  about  8  ft.  or  96  ins.,  which  can  be  determined  accu- 
rately enough  by  scale.  Then  substituting  in  Formula  Q,  Art.  73,  we 
have 

Qfi 
16,000-70^=9,700  Ibs. 

for  the  allowable  unit  stress. 

Dividing  the  greatest  compressive  stress,  which  occurs  in  the  lateral 
bl,  by  this  intensity  we  have 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES  223 

for  the  required  cross-section  of  the  lateral,  and  the  assumed  angle  has 
2.48n".  Then,  as  the  ratio  of  L/r  is  only  about  90,  the  assumed  angle  is 
safe  in  the  case  of  the  greatest  compressive  stress. 

As  is  seen  above  the  greatest  tension  is  18,100  Ibs.,  which  occurs  also 
in  lateral  bl.  Then  we  have 

18,100 
lMOO=1-138q-m8- 

for  the  required  net  cross-section,  and  as  the  angle  assumed  has 
[2.48  -(f"xl")]  2.11n"  it  is  quite  safe  for  the  greatest  tension.  So 
we  will  use  1  —  L  3-J"  x  3^"  x  f  "  for  each  end  lateral.  But  as  this  angle 
is  the  smallest  permitted  in  this  work  the  laterals  throughout  will  each 
be  made  of  1—  3J"  x  3  J"  x  f"  angle. 

The  intermediate  cross-frames  are  not  subject  to  theoretical  analyses 
and  hence  their  design  is  governed  principally  by  experience  and  sense  of 
fitness. 

The  end  cross-frames  can  be  fairly  well  analyzed.  The  stress  in  the 
top  angle  of  the  frame  can  be  taken  equal  to  one-half  of  the  lateral  force 
per  foot  of  span  multiplied  by  one-half  the  length  of  the  span,  and  this 
force  multiplied  by  the  secant  of  the  slope  of  the  diagonals  in  the  frame 
is  equal  to  the  stress  in  each  of  these  diagonals.  The  bottom  angle  of  the 
frame  has  no  stress  (theoretically). 

Then  according  to  the  above  we  have 

-^2-x  25  =  8,750  Ibs. 


for  the  stress  in  the  top  angle  of  the  end  frame,  and  as  the  diagonals  of 
the  frame  slope  about  45°  with  the  horizontal  we  have 

8,750x1.4  =  12,250  Ibs. 

for  the  stress  in  the  diagonals  of  the  end  frame.  From  this  it  is  seen  that 
very  small  angles  are  required  theoretically  for  the  end  frames  which 
are  subjected  to  greater  stresses  (apparently)  than  the  intermediate 
frames,  but  as  3J"  x  3|"  x  f  "  angles  are  practically  the  minimum  size 
used  in  railroad  bridges  we  will  use  this  size  in  all  of  the  frames. 

This  completes  the  necessary  preliminary  calculations  except  for  the 
preliminary  estimate  of  the  dead  weight.  The  stress  sheet  for  the  span, 
as  shown  in  Fig.  169,  can  be  drawn  from  the  information  given  in  the 
above  calculations.  The  estimate  of  the  dead  weight  in  this  case  will  be 
deferred  until  after  the  detail  shop  drawing  of  the  span  is  considered  so 
as  to  familiarize  the  student  with  details  before  taking  up  preliminary 
estimates  of  dead  weight. 

136.  Making  of  the  Shop  Drawing.  —  After  the  stress  sheet 
(Fig.  169)  is  completed  the  shop  drawing  for  the  span,  as  shown  in  Fig. 
171,  can  be  made.  This  work  is  known  as  detailing  and  includes  not  only 
the  drawing  but  the  calculations  for  the  details  as  well. 

To  evolve  this  drawing  (Fig.  171)  the  details  are  first  drawn  in 
pencil  to  a  f  "  scale  upon  a  24"  x  36"  sheet  of  ordinary  drawing  paper 
After  selecting  the  relative  positions  for  the  different  views,  so  that  no 
part  of  the  sheet  will  be  crowded,  we  start  the  drawing,  as  shown  in  Fig. 
170,  by  drawing  the  center  line  CC  of  the  span.  Then  scaling  off  25  ft., 


224 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES 


225 


the  line  BB  through  the  center  of  bearing  is  drawn.  Next,  the  line  EE, 
through  the  end  of  the  span,  must  be  located  and  drawn.  The  end 
stiffeners  should  be  placed  6"  back  to  back  as  shown,  this  being  about  the 
minimum  distance  permissible  on  account  of  driving  rivets  in  the  out- 
standing legs  of  the  stiffeners  connecting  to  the  end  cross-frames.  Then 
allowing,  say,  2"  from  the  end  pair  of  stiffeners  to  the  end  of  the  girders, 
we  have  3"  +  3J"  +  2"  =  8 \"  for  the  distance  from  the  line  BB  to  line 


•?->  •-    ^a--te-^&«aiE<^^=^^ss~j^ 


Fig.   170 

EE,  and  the  line  EE  locating  the  end  of  the  span  can  then  be  drawn. 
Then  the  top  plan  of  each  girder  is  drawn  as  shown  in  Fig.  170.  The 
next  thing  after  that  is  to  locate  the  cross-frames  as  shown,  so  that  the 
lateral  bracing  will  be  symmetrical  about  the  center  line  CC,  and  the 
laterals  will  all  have  the  same  length  (thus  saving  templets).  Now  there 
will  be  a  pair  of  stiffeners  on  each  girder  at  each  cross-frame,  one  of 
which,  in  each  case,  will  be  connected  to  a  frame,  and  when  the  position 
of  the  cross-frames  is  fixed  these  stiffeners  to  which  the  frames  connect 
are  also  fixed  in  position.  Then  the  elevation  of  one  of  the  girders, 
always  the  far  girder,  can  be  drawn  and  the  drawing  completed  as  far  as 
is  shown  in  Fig.  170. 

The  next  thing  is  to  draw  the  laterals  and  the  lateral  plates  connect- 
ing the  laterals  to  the  girders  (as  shown  in  Fig.  171).  As  the  laterals 
are  all  of  the  same  section  and  the  strength  of  each  is  greater  than  is 
required  by  the  stress,  each  end  connection  must  contain  enough  rivets 
to  develop  the  strength  of  the  lateral  in  compression  (see  specifications). 
Now,  as  shown  above,  each  lateral  will  safely  resist  9,700  Ibs.  per  square 
inch  of  cross-section.  Then,  as  2.48n"  is  the  area  of  cross-section  of 
each  lateral,  we  have  2.48x9,700  =  24,000  Ibs.  for  the  total  allowable 
compressive  stress  in  each  lateral.  Then  using  £•"  field  rivets  for  the 


226 


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233 


234  STRUCTURAL  ENGINEERING 

lateral  connections,  the  value  of  which  in  single  shear  will  govern,  6,000 
Ibs.  being  the  allowable  value  of  each,  we  have 

24,000  _ 
6,000  = 

for  the  number  of  rivets  required  in  each  end  of  each  lateral. 

Now,  the  number  of  rivets  connecting  the  lateral  plates  to  the  girders 
must  be  sufficient  to  take  the  component  along  the  girder  of  the  lateral  or 
laterals.  Twenty-four  thousand  pounds  is  the  strength  of  each  lateral, 
and  using  J"  field  rivets  in  single  shear,  the  number  of  rivets  for  inter- 
mediate points  will  be 

24,000  xsinA       / 24,000  x  .695 


6,000       / 

The  number  used  would  be  not  less  than  6,  or  3  for  each  lateral. 

The  laterals  and  lateral  plates  can  now  be  drawn  in  the  top  plan 
(shown  in  Fig.  171),  giving  the  spacing  of  the  rivets  at  each  connection, 
and  determining  clearances  and  the  sizes  of  plates  by  scale  as  the  work 
progresses,  using  separate  sketches  for  each  point  drawn,  say,  to  1-J"  or 
3"  scale.  After  this  work  is  completed  the  stiffeners  between  the  cross- 
frames  can  be  drawn  on  the  elevation  of  the  girder,  care  being  taken  to 
space  the  stiffeners  so  as  to  miss  the  lateral  plates  as  much  as  possible. 
Then  the  longitudinal  section  showing  the  bottom  flange  can  be  drawn. 

After  this  the  next  thing  is  to  space  the  rivets  in  the  vertical  legs  of 
the  flange  angles.  As  the  live  load  is  applied  directly  to  the  top  flanges 
and  the  specifications  ignore  the  one-eighth  of  the  web  in  determining  the 
spacing  of  flange  rivets,  the  problem  of  determining  the  spacing  comes 
under  Case  III,  Art.  116. 

We  assume  that  each  wheel  of  the  live  load  is  distributed  over  three 
ties  and  assuming  ties  8"  wide  and  spaced  6"  apart  (face  to  face)  we 
have  42"  for  the  length  over  which  each  wheel  is  distributed.  Then  for 
the  greatest  vertical  load  per  linear  inch  of  flange  due  to  the  live  load 
(Cooper's  £50),  when  the  maximum  shear  occurs,  we  have 

=  595  Ibs.;  say  600  Ibs., 

•±£/ 

per  linear  inch,  and  adding  a  100  per  cent  for  impact  will  make  a  total 
of  1,200  Ibs.  per  inch. 

Now  referring  to  Case  III,  Art.  116,  we  have 

v=   1,200  Ibs. 
R=    7,880  Ibs.,  which  is  the  allowable  bearing  of  a   J"  shop 

rivet  on  the  f "  web. 
S=   216,700  Ibs.,  which  is  the  maximum  shear  at  either  end  of 

each  girder. 
h=    66  ins. 

Then  substituting  these  values  in  Formula  (4),  Art.  116,  we  have 

7  880 
p=     .  =2.27  ins.  (about) 

/  216,700  \2 

^/1,2002    ' 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


235 


for  the  theoretical  spacing  of  the  rivets  near  the  ends  of  the  girders,  and 
the  theoretical  spacing  at  other  points  along  the  girders  can  be  deter- 
mined by  computing  the  maximum  shear  at  each  of  the  points  and  apply- 
ing Formula  (4),  Art.  116,  in  each  case.  However,  the  following  graph- 
ical method  of  determining  the  spacing  is  preferable  as  it  saves  time  and 
is  accurate  enough  for  practical  work: 

The  shear  due  to  dead  load  is  quite  small  compared  with  that  due  to 
live  load  and  impact,  so  that  the  maximum  shear  can  be  assumed  to  vary 
across  the  span  as  the  ordinates  to  a  parabola  (see  Art.  55). 

Now  the  above  spacing  (2.27")  gives  about  5.3  rivets  per  foot  of 
girder.  Then  if  we  lay  off  a  vertical  line  AB  (Fig.  172)  equal  to  5.3", 
and  taking  a  pair  of  dividers  divide  the  line  AB  into  any  convenient 


Fig.   172 

number  of  equal  parts  and  from  A  step  off  the  same  number  of  equal 
parts  on  the  line  AC  (the  equal  parts  on  the  line  AC  can  be  taken  any 
length,  just  so  A  to  C  is  a  convenient  distance),  and  construct  the 
parabola  CDB  (see  Art.  58)  and  draw  the  horizontal  lines  to  the  curve 
through  the  points  1,  2,  3,  etc.,  we  have  a  diagram  from  which  the 
required  spacing  of  the  rivets  at  any  point  between  the  end  and  the 
center  of  the  span  can  be  obtained,  as  the  distance  AC  corresponds  to 
the  length  of  the  span  and  the  line  AB  gives  the  shear  (correspondingly) 
in  number  of  rivets  per  foot  of  girder.  From  the  diagram  (Fig.  172)  it 
is  seen  that  about  one  and  one-third  rivets  per  foot  are  required  at  the 
center  of  the  span,  and  practically  three  rivets  per  foot  at  the  quarter 
point.  The  required  spacing  at  all  other  points  from  A  to  the  center  of 
the  span  is  seen  at  a  glance. 

It  is  not  practical,  if  not  impossible,  to  space  the  rivets  throughout 
a  girder  according  to  the  theoretical  requirements,  as  no  two  spaces 
would  be  the  same  (which  is  impractical)  and  the  spaces  next  to  stif- 
feners  can  not  be  less  than  about  8J",  which  in  some  cases  is  greater  than 
the  required;  so  the  best  we  can  do  is  to  fit  the  rivets  in  between  the 
stiffeners  to  about  the  theoretical  spacing. 


236  STRUCTURAL  ENGINEERING* 

The  theoretical  spacing  of  the  flange  rivets  as  determined  in  the 
manner  shown  above  will  satisfy  the  requirements  of  the  specifications, 
which  are  as  follows: 

"The  flanges  of  plate  girders  shall  be  connected  to  the  web  with 
sufficient  number  of  rivets  to  transfer  the  total  shear  at  any  point  in  a 
distance  equal  to  the  effective  depth  of  the  girder  at  that  point  combined 
with  any  load  that  is  applied  directly  to  the  flange." 

This  is  simply  a  practical  manner  of  obtaining  approximately  the 
theoretical  spacing.  To  show  the  justification  of  the  above  requirement 
given  in  the  specifications  let  AB,  Fig.  173,  represent  a  plate  girder. 
Let  M  and  S  be  the  bending  moment  and  shear,  respectively,  at  section 
C.  Then  for  the  bending  moment  at  section  D  we  have 

M'  =  M  +  Sx-2Pz.     (See  Art.  66.) 

Now,  Sx  -  2  Pz  is  the  difference  between  the  moments  at  the  two  sections. 
If  the  intervening  loads  be  neglected,  which  is  an  error  on  the  side  of 

safety,  the  difference  between  the  two 
moments  is  Sx,  and  if  this  be  divided  by 
the  effective  depth  the  result  would  be 
the  difference  between  the  flange  stresses 
at  the  two  sections,  which  would  be 
equal  to  the  shear  when  x  is  equal  to  the 
Fig.  173  effective  depth,  as  is  readily  seen.  This 

difference  of  flange  stress  is  simply  the 

increment  of  the  flange  stress  between  the  two  sections,  and  of  course 
there  should  be  a  sufficient  number  of  rivets  to  transfer  this  from  the 
web  to  the  flange  and  at  the  same  time  support  whatever  vertical  load 
there  is  applied  to  the  flange.  This  is  entirely  in  accord  with  Arts.  115 
and  116. 

Now,  having  the  stiffeners  spaced  as  shown  in  Fig.  171  we  can  begin 
spacing  the  rivets  in  the  vertical  legs  of  the  flange  angles.  Beginning  at 
the  end  of  the  girder,  the  theoretical  spacing  as  given  above  is  2.27",  but 
we  will  use  2J"  in  order  to  have  practical  spaces.  So  we  obtain  the 
spacing  from  the  end  of  the  girder  to  the  first  intermediate  stiffener  as 
shown  in  Fig.  171.  Between  the  first  and  second  stiffeners  we  can 
increase  the  spacing  a  little  as  shown  by  the  diagram  in  Fig.  172,  and 
between  the  second  and  third  stiffeners  we  can  increase  the  spacing  a 
little  more  and  so  on  towards  the  center  of  the  span  until  the  limit  of  6" 
spacing  is  reached,  which  according  to  the  specifications  must  not  be 
exceeded.  It  is  practice  to  limit  the  fractions  in  the  spacing  to  no  less 
than  y.  If  necessary  the  stiffeners  can  be  shifted  slightly  to  suit  the 
spacing. 

After  the  rivets  are  spaced  in  the  vertical  legs  of  the  flange  angles 
the  rivets  in  the  horizontal  legs,  connecting  the  cover  plates  to  the  angles, 
can  be  spaced.  The  theoretical  requirement  in  this  case  is  that  the  rivets 
be  spaced  so  that  the  number  of  rivets  per  foot  will  be  sufficient  to 
transmit  the  part  of  the  flange  increment  taken  by  the  cover  plates.  The 
part  of  the  flange  increment  taken  at  any  point  by  the  cover  plates  will 
be  to  the  total  flange  increment  at  that  point  as  the  area  of  the  cross- 
section  of  the  cover  plates  is  to  the  total  area  of  the  cross-section  of  the 
flange  at  that  point.  As  an  example,  let  M  be  the  bending  moment  found 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  237 

at  section  C  (Fig.  173)  and  AT  the  bending  moment  at  section  D,  which 
is  a  short  distance  to  the  right  of  C  (say,  a  foot).  Then  for  the  total 
flange  increment  between  the  two  sections  we  have 


where  h  is  the  effective  depth  of  the  girder.  Now  it  is  obvious  that  there 
must  be  a  sufficient  number  of  rivets  in  the  vertical  legs  of  the  flange 
angles  between  the  two  sections  to  take  all  of  this  increment  as  the  total 
increment  is  transmitted  thrpugh  these  rivets,  but  as  the  cover  plates 
take  only  their  proportional  part  of  the  increment,  the  number  of  rivets 
connecting  them  to  the  flange  angles  need  be  only  sufficient  to  take  their 
part.  So  if  r  be  the  allowable  stress  on  each  rivet  connecting  the  cover 
plates  to  the  flange  angles,  and  A  the  total  area  of  cross-section  of  the 
flange  between  section  C  and  D,  and  A",  the  area  of  the  cover  plates,  we 
have 

F     A" 
n  =  —  x—r 
r      A 

for  the  number  of  rivets  required  in  the  cover  plates  between  the  two 
sections  and  the  number  of  rivets  required  to  connect  the  cover  plates  to 
the  flange  angles  at  any  other  point  along  the  girder  can  be  determined 
in  the  same  manner,  but  the  same  result  can  be  obtained  in  any  case  with 
much  less  work  by  using  such  a  diagram  as  shown  in  Fig.  172.  So  we 
will  use  the  above  diagram  (Fig.  172)  in  this  case.  Beginning  at  the 
center  of  the  span,  the  number  of  rivets  per  foot  required  to  take  the  total 
flange  increment  is  1J.  Then  as  the  net  area  of  cross-section  of  the 
flange  at  that  point,  not  including  the  one-eighth  of  the  web,  is  21.86n", 
and  the  net  area  of  the  cover  plates  is  11.25n",  we  have 

'      "        '  Hxgff  =  0.68  (about) 

for  the  number  of  rivets  per  foot  in  the  cover  plates,  which  gives  a 
spacing  of  17.6".  This  would  be  the  spacing  if  the  allowable  stress  on 
the  rivets  in  the  horizontal  legs  were  the  same  as  in  the  vertical.  But  as 
the  rivets  in  the  horizontal  legs  must  be  considered  in  single  shear  and 
the  rivets  in  the  vertical  legs  in  bearing  on  the  f  "  web,  the  above  number 
(0.68)  must  be  multiplied  by  7,880/7,200.  So  we  have 

11.25      7,880        _ 


for  the  number  of  rivets  per  foot  in  the  cover  plates  at  the  center  of  the 
span,  which  gives  (12/0.75)  16"  spacing,  but  as-  there  are  two  rows  of 
rivets  in  the  cover  plates  the  spacing  of  the  rivets  in  each  flange  angle 
would  really  be  twice  this,  or  32". 

Again,  from  the  diagram    (Fig.   172)   the  required  number  of  the 
rivets  in  the  vertical  legs  of  the  flange  angles  at  the  quarter  point  is  3  per 
foot.     Then  for  the  number  of  rivets  required  at  that  point  in  the  cover 
plate  (as  the  TV'  plate  only  extends  to  about  that  point),  we  have 
6.00       7,880 


238  STRUCTURAL  ENGINEERING 

which  gives  a  spacing  of  10.1"  or  20.2"  along  each  angle.  Now,  if 
other  points  along  the  girder  be  considered  it  will  likewise  be  found  that 
the  required  spacing  of  the  rivets  in  the  cover  plate  or  plates  will  exceed 
the  6"  maximum  allowed.  So  that  if  we  were  to  space  the  rivets  in  the 
cover  plates,  using  the  6"  maximum  allowable  throughout,  there  would  yet 
be  an  excess  of  rivets.  Now  the  spacing  of  the  rivets  in  the  lateral  plates 
is  determined  when  the  laterals  are  detailed,  as  stated  above,  and  as 
these"  rivets  must  be  spaced  to  suit  those  details  it  only  remains  to  fit  the 
rivets  in  between  the  lateral  connections  using  as  near  the  maximum  6" 
spacing  as  will  fit  in  nicely  without  interfering  with  the  stiffeners.  So 
we  obtain  the  spacing  shown  in  the  plan  of  the  top  flanges  (Fig.  171)  and 
the  same  spacing  can  be  used  in  the  bottom  flange,  as  shown  in  the  longi- 
tudinal section  below  the  girder.  As  is  evident,  the  spacing  selected  at 
the  different  points  depends  entirely  upon  individual  judgment  and  hence 
no  two  persons  are  likely  to  make  the  same  selection.  However,  this  is 
not  a  serious  matter  at  all  as  any  reasonable  variation  is  permissible. 

We  shall  next  space  the  rivets  in  the  stiffeners.  Now  according  to 
shop  practice  the  rivets  in  the  stiffeners  should  all  line  up  horizontally 
all  the  way  across  the  girders. 

There  is  no  definite  rule  for  determining  the  number  of  rivets 
required  in  the  intermediate  stiffeners  other  than  that  there  should  be  a 
sufficient  number  to  clamp  the  stiffeners  firmly  to  the  web,  and  the 
maximum  6"  spacing  usually  suffices  for  this,  but  the  end  stiffeners  should 
be  connected  with  a  sufficient  number  to  transmit  the  total  end  shear 
from  the  web  to  the  masonry,  as  these  stiffeners  are  really  columns  bear- 
ing against  the  bottom  flange  of  the  girder  and  really  transmit  this  force. 
It  is  customary  to  space  the  rivets  in  the  end  stiffeners  so  that  about 
every  other  rivet  of  this  spacing  can  be  omitted  for  the  spacing  in  the 
intermediate  stiffeners  whereby  an  economic  spacing  that  lines  up  hori- 
zontally throughout  the  full  length  of  the  girder  is  obtained.  It  is 
economic  to  crimp  all  stiffeners  on  all  girders  over  three  feet  deep,  but  it 
is  usual  shop  practice  to  put  fillers  under  end  stiffeners  and  under  the 
stiffeners  at  all  points  where  cross-frames  connect.  So  we  shall  put 
fillers  under  the  end  stiffeners  and  under  the  stiffeners  at  cross-frames. 
The  rivets  in  the  stiffeners  are  in  double  shear  and  bearing  on  the  f  "  web. 
Then,  using  J"  shop  rivets,  we  have 

12,000x0.6x2  =  14,400  Ibs. 
for  the  allowable  shear  on  each  rivet  and 

24,000  xjx  1  =  7,880  Ibs. 

for  the  allowable  bearing  on  each  rivet.  So  the  bearing  governs  the 
number  of  rivets. 

Now,  taking  the  case  of  the  end  stiffeners,  the  maximum  end  shear 
being  216,700  Ibs.,  we  have 

216,700 

=27.5,  say,  28, 


for  the  number  of  rivets  required  in  the  two  pairs  of  end  stiffeners  at  each 
support.  But,  as  the  end  stiffeners  are  on  fillers  there  must  be  an  excess 
of  50  per  cent,  according  to  the  specification,  so  42  rivets  will  be  used  as 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  239 

shown  (Fig.  171),  and  by  omitting  every  other  rivet  in  the  spacing  shown 
in  the  end  stiffeners  we  obtain  the  spacing  shown  in  the  intermediate 
stiffeners.  The  rivets  in  the  stiffeners  can  be  made  to  line  up  horizontally 
without  any  trouble  if  3"  spacing  be  used  in  the  end  stiffeners.  In  case 
this  spacing  gives  more  rivets  in  the  end  stiffeners  than  are  required, 
some  of  the  rivets  can  be  omitted,  thus  making  some  6"  spaces. 

The  distance  from  the  toe  of  any  flange  angle  to  the  first  rivet  out 
(in  the  stiffener)  from  the  flange,  according  to  shop  practice,  should  be 
equal  to  not  less  than  twice  the  thickness  of  the  flange  angle  plus  1J". 
So  in  this  case  we  have  1  J"  +  1J"  +  1-J"  =  4-J"  for  the  minimum  allowable 
distance  from  the  outer  gauge  line  in  the  flange  angles  to  the  first  rivet 
out  from  the  flange.  These  distances  are  given  in  the  spacing  at  the  end 
of  the  girder  (Fig.  171)  as  4-f",  which  exceeds  the  minimum  allowed 
by  \" .  These  spaces  should  not  be  less  in  any  case  nor  much  more  than 
J"  greater  than  the  minimum  allowed  in  conformity  with  the  practice 
mentioned  above. 

After  the  rivets  are  spaced  in  the  stiffeners  the  web  splices  can  be 
calculated  and  drawn  to  conform  to  this  spacing.  The  number  of  web 
splices  is  always  limited  in  all  cases  to  as  few  as  is  possible,  which 
depends  altogether  upon  the  maximum  length  of  the  web  plates  obtainable 
from  the  steel  mills.  Each  splice  should  be  at  a  stiffener.  The  maximum 
length  of  72"  x  f"  plates  is  given  in  Table  9  as  30  ft.  So  one  splice  is 
necessary  in  the  above  50-ft.  span. 

Let  us  use  the  type  of  splice  shown  at  (a),  Fig.  150,  Art.  117.  The 
first  thing  to  do  is  to  determine  the  size  of  the  longitudinal  splice  plates 
next  to  the  flanges.  These  plates  should  have  at  least  three  horizontal 
rows  or  rivets  in  them  in  order  to  get  a  good  hold  on  the  web.  This  much 
is  simply  a  matter  of  judgment.  So  let  us  assume  that  there  are  three 
horizontal  rows  of  rivets  in  each  pair,  then  the  distance  from  the  center 
of  gravity  of  the  rivets  in  the  top  pair  of  plates  to  the  center  of  gravity 
of  the  rivets  in  the  bottom  pair  is  equal  to  the  distance  from  the  second 
rivet,  from  the  top  flange,  to  the  second  rivet  from  the  bottom  flange, 
which,  as  seen  from  the  spacing  on  the  girder  (Fig.  171)  is  4  ft.  or  48". 
Now,  according  to  Art.  117,  using  the  figures  given  above,  we  have 


from  which  we  obtain 

F  =  79,600  Ibs. 

for  the  direct  longitudinal  stress  on  each  pair  of  plates. 

Then  we  have  79,600  -H  (16,000)  48/71  =  7.35n"  for  the  required  net 
area  of  each  pair  of  plates,  or  about  3.67n"  for  each  plate.  Allowing  \" 
clearance  between  these  plates  and  the  flange  angles  and  the  same  clear- 
ance between  them  and  the  vertical  splice  plates,  we  obtain  10J"  for 
their  width.  Now,  let  us  assume  that  they  are  \"  thick,  and  that  their 
cross-section  is  reduced  by  three  rivet  holes  (for  J"  rivets),  then  we  have 


for  the  net  area  of  cross-section  of  each  plate,  which  is  about  equal  to  the 
required  area.     But  if  the  plates  be  \"  thick  a  -fa"  filler  would  be 


240  STRUCTURAL  ENGINEERING 

required  under  each  of  the  stiffeners  at  the  splice  and  to  avoid  these  thin 
fillers  we  will  make  the  plates  as  thick  as  the  flange  angles,  which  is  iV'- 
This,  of  course,  gives  a  little  excess  area,  but  the  result  obtained  justifies 
its  use. 

The  next  thing  is  to  determine  the  number  of  rivets  in  these  longi- 
tudinal plates.  As  the  number  of  rivets  depends  upon  their  allowable 
bearing  upon  the  f "  web,  which  is 

(7,880)  M  =  5,330  Ibs., 


we  have 

79,600 


=  15  (about) 


5,330 

for  the  number  of  rivets  required  on  each  side  of  the  splice  in  each  pair  of 
plates ;  and  spacing  them  in  the  plates,  so  as  to  conform  to  the  spacing  in 
the  stiffeners  and  flange  angles  previously  established,  we  obtain  the 
spacing  shown  in  Fig.  171. 

The  next  thing  is  to  determine  the  size  of  the  vertical  splice  plates 
and  the  number  of  rivets  required  in  them  to  transmit  the  maximum  shear 
at  the  splice  as  explained  in  Art.  117. 

/' 


,-  ci 


& 


so' 


a 


Fig.  174 


Placing  the  wheel  loads  on  the  span  so  that  wheel  2  will  be  at  the 
splice,  as  shown  in  Fig.  174,  and  taking  moments  about  the  support  B 
(using  Table  A),  we  have 


50 
for  the  reaction  at  A,  then,  we  have 

(34,900-10,000)^=31,200  Ibs. 

for  the  maximum  live-load  shear  at  the  splice. 
For  the  impact  we  have 


Now,  as  the  dead-load  shear  at  the  splice  is  zero,  we  have 
31,200  +  28,800  =  60,000  Ibs. 

for  the  total  maximum  shear  at  the  splice. 
Then  we  have 

60,OOP. 

IoW  =  6'0s*ins- 

for  the  required  net  area  of  the  vertical  section  of  the  vertical  splice 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 

plates,  but  as  the  splice  plates  are  made  •£$"  thick  in  order  to  avoid  the 
•£Q"  fillers  under  the  stiffeners,  referred  to  above,  there  will  be  consider- 
able  more  metal  in  them  than  is  required  for  shear,  so  they  will  be  amply 
strong  as  far  as  shear  is  concerned.  This  is  practically  always  the  case, 
for  the  splice  plates  cannot,  according  to  the  specifications,  be  less  than 
f"  in  thickness  and  the  two  of  them  will  always  be  thicker  than  the  web 
they  splice.  The  number  of  rivets  required  to  connect  these  plates  to  the 
web  will  depend  upon  their  allowable  bearing  on  the  f"  web  as  shown 
above.  So  we  have 

60,000     „„ 

=7.6,  say,  8, 


for  the  required  number  of  rivets  on  each  side  of  the  splice,  but  a  greater 
number  will  be  used  as  there  must  be  two  rows  of  rivets  on  each  side  of 
the  splice  in  order  to  securely  clamp  the  plates  to  the  web  and  the  spacing 
must  conform  to  that  already  established  in  the  stiffeners.  So  following 
out  these  requirements  without  spacing  the  rivets  farther  apart  than  6", 
we  obtain  the  spacing  shown  in  Fig.  171. 

Now  to  test  the  splice  as  a  whole,  let  S  =  the  horizontal  stress  on  each 
of  the  rivets  in  the  horizontal  rows  farthest  out  from  the  center  of  the 
web,  that  is,  in  the  rows  next  to  the  flanges,  and  let  V  =  the  vertical  shear 
on  each  rivet,  which  we  will  assume  to  be  the  same  for  each  rivet  in  the 
splice,  that  is, 


Now  taking  moments  about  the  center  of  the  web,  as  explained  in 
Art.  117,  we  have 


16,000x3.37x71, 
from  which  we  obtain 

5  =  5,118  Ibs. 

Then  for  the  maximum  resultant  stress  which  occurs  on  the  outer 
rivets,  we  have 

^  =  V/5,1182  +  1,764:2  =  5,413  (about), 

which  is  about  567  Ibs.  less  than  the  allowable,  which  is  5,980  =  (7,880) 
27x2/71.  So  the  splice  is  all  right  and  the  drawing  of  it  can  be  com- 
pleted as  shown. 

It  is  seen  from  the  above  equation  that  the  rivets  near  the  center  of 
the  web  are  of  little  value,  in  resisting  bending. 

We  will  next  take  up  the  detailing  of  the  cross-frames.  The  first 
thing  to  do  is  to  draw  the  vertical  lines  representing  the  centers  of  the 
girders  as  shown  in  the  detail  to  the  right  of  the  girder.  Next  the  top 
and  bottom  horizontal  angles  should  be  located  and  the  position  of  the 
diagonal  determined.  Then  the  next  thing  is  to  determine  the  size  of 
each  of  the  connection  plates  marked  ok.  The  size  of  each  of  these  plates 
will  depend  upon  the  number  of  rivets  in  the  ends  of  the  diagonals.  As 
the  diagonals  take  both  tension  and  compression  we  consider  them  as 


242  STRUCTURAL  ENGINEERING 

compression  numbers,  and  assuming  them  independent  of  each  other,  each 
has  a  length  of  97"  (about).     Then  we  have 

07 
16,000-70-^=9,650  Ibs. 

for  the  allowable  unit  compressive  stress  on  each  diagonal, 
and  9,650  x  2.48  =  23,900  Ibs. 

for  the  allowable  stress  on  each.     Then  using  J"  shop  rivets,  we  have 

23,900 


7,220 


=  3.3,  say,  4, 


for  the  number  of  rivets  required  in  each  end  of  the  diagonals. 

Now  a  large  scale  drawing  of  one  of  the  four  plates  ak,  showing  the 
entire  detail  at  that  point  (girder  and  all)  can  be  made  from  which  the 
location  of  the  rivets  and  all  necessary  clearances  can  be  determined. 
Then  the  cross-frame  and  the  cross-section  of  the  girder  can  be  drawn  as 
shown  and  the  pencil  drawing  is  then  complete  and  the  tracing  of  the 
same  can  be  made  and  by  adding  the  required  list  and  writing  on  the 
title  and  general  notes  we  have  the  shop  drawing  for  the  span  completed 
except  for  the  anchor  bolts  and  cast  pedestals  which  are  detailed  on 
sketch  sheets  included  in  the  above  shop  bills  which  are  made  after 
the  shop  drawing  is  completed.  After  the  shop  drawing  and  bills  are 
checked  and  approved  blueprints  are  made  of  them  which  are  used  in  the 
fabricating,  shipping,  and  erecting  of  the  span. 

137.  Estimate  of  Dead  Weight  and  Cost  of  Span. — From  the 
shop  drawing  (Fig.  171)  we  have  the  following  weight  of  metal: 

Estimate  of  the  weight  of  main  girders. 

l_web  72  x  f  x  91.8  Ibs.  x  51.4' 4,718  Ibs. 

4— Ls  6x6x^x21.9  Ibs.  x  51.4' 4,503  Ibs. 

2— cov.  pis.  14  xTVx  20.82  Ibs.  x  27.0'  (top  and  bot.) 1,124  Ibs. 

1— cov.  pi.  14  x  i  x  23.80  Ibs.  x  51.4'  (top) 1,223  Ibs. 

1— cov.  pi.  14  x  4  x  23.80  Ibs.  x  37.0'  (bot.) 881  Ibs. 

8— Ls  5  x  3i  x  J  x  13.6  Ibs.  x  6.0'  (end  stiff.) 653  Ibs. 

26— Ls  5  x  3£  x  |  x  10.4  Ibs.  x  6.0'  (int.  stiff.) 1,622  Ibs. 

12— fills.  3JxT%x6.7  Ibs.  x  5.0' 402  Ibs. 

2— sp.  pis.  16  x  T%  x  30.6  Ibs.  x  3.2' 196  Ibs. 

4— sp.  pis.  10xT%x  19.14  Ibs.  x  2.1' 161  Ibs. 

2— sole  pis.  14xf  x35.71  Ibs.  x  1.3' . 93  Ibs. 

15,576  Ibs. 

1,015*  rivets  @  .44  Ibs.  per  pair  of  heads  (  J"  rivets) 446  Ibs. 

Total  weight  of  one  main  girder ;  .16,022  Ibs. 

Total  weight  of  two  main  girders 32,044  Ibs. 

Total  weight  of  main  girders  per  foot  of  span  equals 

=641  Ibs. 


*For  weight  of  rivet  heads  see  Carnegie  Handbook. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  243 

It  Till  be  seen  from  the  above  that  the  weight  of  the  rivet  heads  is 
about  3  per  cent  of  the  weight  of  the  other  material  in  the  girders. 

Estimate  of  the  weight  of  laterals  and  lateral  plates. 

8— Ls  3J  x  31  x  f  x  8.5  Ibs.  x  8.2' 558  Ibs. 

7— pis.  12  x  f  x  15.3  Ibs.  x  2.6' 278  Ibs. 

2— pis.  12  x  §  x  15.3  Ibs.  x  2.0' 61  Ibs. 

2— pis.  12  x  f  x  15.3  Ibs.  x  1.3' 40  Ibs. 

9— pis.  12  x  f  x  15.3  Ibs.  x  1.0' 137  Ibs. 

4— pis.  12xf  x  15.3  Ibs.  x  0.8' 49  Ibs. 

Total  weight  of  laterals  and  lateral  plates 1,123  Ibs. 

Estimate  of  the  weight  of  cross-frames. 

2— Ls  3^  x  3J  x  f  x  8.5  Ibs.  x  6.2' 105  Ibs. 

2— Ls  3|  x  31  x  f  x  8.5  Ibs.  x  7.3' 124  Ibs. 

4— pis.  14  x  f  x  17.86  Ibs.  x  1.1' 78  Ibs. 

1— pi.  8J  x  f  x  10.84  Ibs.  x  0.8' 9  Ibs. 

316  Ibs. 
33  rivets  @  0.44  Ibs .     14  Ibs. 

Total  weight  of  one  frame 330  Ibs. 

Total  weight  of  five  frames 1,650  Ibs. 

Total  weight  of  lateral  system  =  1,123  +  1,650  =  2,773  Ibs. 

p    lyiVQ 

Total  weight  of  lateral  system  per  foot  =—77; —  =  55  Ibs.  per  ft.  of  span. 

oO 

Estimate  of  the  total  effective  dead  load  per  ft.  of  span. 

2  girders 641  Ibs. 

Lateral  system 55  Ibs. 

Deck   (track) 400  Ibs. 

Total 1,096  Ibs. 

Difference  between  the  assumed  dead  load  and  estimated  =  1,150  - 
1,096  =  54  Ibs.,  which  is  less  than  10  per  cent,  so  no  recalculation  is 
necessary  on  account  of  error  in  assumed  dead  load. 

Summary  of  total  weight  of  metal  in  span. 

2  girders , 32,044  Ibs. 

Lateral  system 2,773  Ibs. 

4  pedestals  @  360  Ibs 1,440  Ibs. 

Anchor  bolts  and  nuts 61  Ibs. 

Total   36,318  Ibs. 

Approximate  Cost  of  Span  Erected:  36,318  Ibs.  @  3^  =  $1,180. 
Three  and  one-quarter  cents  is  only  a  fair  average  price  for  this  class  of 
work  erected.  The  price  will  vary  from  2^  to  4^.  It  depends  upon  the 
market  price  of  steel  and  the  freight. 


244  STRUCTURAL  ENGINEERING 

It  will  be  seen  from  the  above  tli.it  the  weight  of  rivet  heads  (=  1,000 
Ibs.,  approximately)  is  equal  to  about  3  per  cent  of  the  weight  of  the 
other  metal  (=32,700  Ibs.)  in  the  span,  not  considering  the  pedestals.  So 
in  making  preliminary  estimates  of  such  structures  the  weight  of  the  rivet 
heads  can  be  taken  at  about  3  per  cent  of  the  wreight  of  the  other  metal. 

138.  Hints  Regarding  Shop  Bills, —  The  above  shop  bills  are  for 
the  most  part  self-explanatory.  Pages  Nos.  1  and  2  are  for  the  structural 
material  proper.  The  finished  material  is  given  on  the  material  side  of 
the  bills  exactly  as  it  appears  on  the  shop  drawing.  The  length  of 
material  as  obtained  from  the  rolling  mills  is  likely  to  vary  slightly  from 
the  length  ordered,  so  in  case  of  long  pieces  where  an  under  run  would  be 
objectionable  a  \"  or  so  is  added  in  the  length  given  on  the  mill  order  side 
as  shown.  In  case  of  short  pieces  having  a  length  of  10  ft.  and  under, 
and  having  the  same  section,  they  are  ordered  in  multiple,  that  is,  they 
are  combined  and  ordered  as  one  or  more  pieces,  as  is  seen. 

All  pieces  planed  (marked  fin.)  on  one  side  are  ordered  -j\r''  thicker 
than  the  finished  thickness  and  J"  thicker  if  planed  on  two  opposite  sides. 
All  pieces  planed  on  the  ends  are  ordered  \"  to  \"  longer  than  the 
finished  length.  This  is  in  addition  to  the  amount  added  for  under  runs. 

The  dimensions  appearing  on  the  mill  order  side  are  only  those 
differing  from  the  finished  dimensions.  In  cases  where  the  material  is  to 
be  ordered  exactly  as  the  finished  material,  it  is  not  repeated  on  the  mill 
order  side,  as  a  rule. 

The  cast-steel  pedestals  are  detailed  on  page  No.  4.  The  required 
area  of  bearing  on  the  masonry  of  each  pedestal  is  given  in  Art.  134  as 
361n".  The  actual  area  is 

20^x18  =  364.5  sq.  ins., 

which  is  about  the  correct  area.  The  thickness  of  metal  in  such  pedestals 
should  not  be  less  than  1J"  in  order  to  insure  the  molds  to  properly  fill. 
The  top  of  each  pedestal  (where  the  girder  rests)  should  be  as  small  as  is 
consistent  with  good  details. 

Everything  shown  on  such  bills  as  the  above  is,  as  a  rule,  written  on 
by  the  draughtsman  making  the  shop  drawing.  The  item  numbers  are 
given  by  the  order  clerk  as  the  final  mill  order  blanks  are  made  out  in  the 
order  department,  which,  as  a  rule,  is  a  separate  department  from  the 
drawing  room.  The  bills  not  mentioned  are  considered  as  being  entirely 
self-explanatory. 


Partial  Design  of  a  75-Ft.  Single-Track  Deck  Plate  Girder  Span 

139.  Data.— 

Length  =  75'  0"  c.c.  end  bearings. 

Dead  Load  =  12  x  75  +  550  =  1,450  Ibs.  per  ft.  of  span. 

Live  Load,  Cooper's  £50. 

Specifications,  A.  R.  E.  Ass'n. 

140.  Calculations. — In  a  manner  similar  to  that  shown  in  Art.  134 
for  the  50-ft.  span,  we  obtain  the  following  for  the  75-ft.  span: 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


245 


Maximum  End  Shear: 
D=    27,200  Ibs. 
L  =  147,200  Ibs. 
7=117,700  Ibs. 

292,100  Ibs. 


Maximum  Moment: 

/)  =    6,117,000  in.  Ibs. 

L  =  28,875,000  in.  Ibs. 

7  =  23,100,000  in.  Ibs. 


58,092,000  in.  Ibs. 


Assuming  the  web  to  be  -£$  in.  thick  we  have 


,-  1055 


/  58,092, 

VAX  16, 


,000 
000 


for  the  economic  depth.  So  we  will  try  a  96  x  •£$"  web.  Using  the 
ordinary  flange  the  effective  depth  is  about  95"  (merely  an  assumption). 
Then  we  have  58,092,000  +  95  =  611,500  Ibs.  for  the  flange  stress,  and 
611,500  -r  16,000  =  38.2n"  for  the  net  area  of  the  flange.  As  stated 
before,  about  half  of  this  area  (minus  J  of  the  area  of  the  web)  should 
be  in  each  pair  of  flange  angles.  This  would  require  angles  which  would 
be  entirely  too  thick  if  6"  x  6"  angles  be  used.  So  either  6"  x  6"  angles 
with  side  plates  and  cover  plates,  or  8"  x  8"  angles  with  cover  plates 
should  be  used.  The  last-mentioned  section  is  really  the  ordinary  flange 
wherein  the  angles  are  8"  x  8". 

Let  us  first  try  the  flange  made  up  of  6"  x  6"  angles,  side  plates 
and  cover  plates.  As  the  effective  depth  in  this  case  is  less  than  in  the 
case  of  ordinary  flanges,  the  area  of  the  flange  will  be  a  little  larger  than 
indicated  above.  So  let  us  assume  the  following  section: 


2— Ls  6"  x  6"  x  f" 
2 — Side  plates  13"  x 
1— Cover  plate  16"  x 
1— Cover  plate  16"  x 
•J  of  web 


Gross  Area  Net  Area 

=  14.22n"-2.5  n"  =  11.72°" 
=  13.00n"-3.0  n"  =  io.00n" 
=  8.00n"  -1.0  n"=  7.00n" 
=  7.00n"-0.87n"=  6.13n" 
=  5.25n" 


42.22n" 


40.10n" 


The  flange  is  assembled  as  shown  in  Fig.  175. 

By  taking  moments  about  the  center  of  the  top  cover  plate  we  obtain 


-_  8x1  +  14.22x2.48  +  13x7^ 
42.22 

for  the  distance  from  the  center  of  the  top 
cover  plate  to  the  center  of  gravity  of  the 
flange.  Then  we  have 

3.24 -0.75  =  2.49  ins. 

for  the  distance  from  the  back  of  the  flange 
angles  to  the  center  of  gravity  of  the 
flange,  and 

96.25-4.9S  =  91.27ins. 
for  the  effective  depth. 


=  3.24  ins. 


Fig.   175 


246  STRUCTURAL  ENGINEERING 

Then  using  this  effective  depth  we  have 

58,092,000 


91.27 
for  flange  stress,  and 


=  637,000  Ibs.  (about) 


637,000      __  _       . 

=  39.8  sq.  ins. 


16,000 

for  the   required  net  area  of  each   flange.      This   shows   that  the  above 
assumed  flange  is  about  correct. 

Next  let  us  try  a  flange  made  up  of  8"  x  8'"  angles  and  cover  plates. 

First  assume  the  following  section: 

Gross  Area  Net  Area 

2_Ls  g"  x  8"  x  ii"  =  21.06n"  -  2.75n"  -  18.31°" 

2—  cov.  pis.  18"  x  J"  =  18.00n"  -  2.00n"  =  16.00n" 

J  of  web  _  =    5.25n" 

39.06n"  39.56n" 

Taking  moments  about  the  center  of  the  top  cover  plate  we  obtain 


jj 

for  the  distance  from  the  center  of  the  top  cover  plate  to  the  center  of 
gravity  of  the  flange,  and  hence  we  have 

1.73  -  0.75  =  0.98  ins.,  say  1  in., 

for  the  distance  from  the  back  of  the  angles  to  the  center  of  gravity  of 
the  flange.  This  gives  96.25  -  2  =  94.25"  for  the  effective  depth.  Then 
we  have 

58,092,000 
94.25  x  16,000  =  38'55sq'm 

for  the  required  net  area  of  each  flange.  This  shows  that  the  section 
assumed  above  is  a  little  too  large.  By  making  one  of  the  cover  plates 
TV'  thick  (instead  of  £")  the  section  will  be  about  correct. 

It  is  seen  from  the  above  that  the  flange  made  up  of  8"  x  8"  angles 
and  cover  plates  is  more  economic  than  the  flange  made  up  of  6"  x  6" 
angles,  side  plates,  and  cover  plates,  and  hence  apparently  should  be 
used.  However,  this  will  depend  mostly  upon  whether  the  8"  x  8"  angles 
can  be  readily  obtained  from  the  rolling  mills  at  the  same  price  as  the 
other  sections.  In  case  several  spans  are  required  it  would  very  likely 
pay  to  use  8"  x  8"  angles,  but  in  case  of  only  one  span  it  would  likely 
be  best  to  use  the  6"  x  6"  angles,  side  plates,  and  cover  plates. 

The  calculations  for  the  remainder  of  this  span  would  be  quite 
similar  to  that  shown  above  for  the  50-ft.  span,  and  when  completed  the 
corresponding  stress  sheet,  detail  shop  drawing,  and  shop  bills  could  be 
made.  Fig.  176  shows  a  partial  detail  of  the  girder  where  the  flanges 
are  made  up  of  6"  x  6"  angles,  side  plates,  and  cover  plates. 

The  student  will  have  no  trouble  in  designing  and  detailing  this  type 
of  girder  if  the  outline  given  above  for  the  50-ft,  span  be  followed. 


Or'OO  O-i;  O  O  O-O-e-O  Q  O  O 0:^ 


I 


\  «»H  tT"n 

/-Cov.P/.  /£x/i 


e— 


-e — © — o— e — e- 


e—  e  —  e< 


Med  holes  £&*/;>  "M/s  end 


/  Tap  bo/ fa  topped /'nto  so 


Fig.   176 


247 


248 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  249 

Partial  Design  of  a  90-Ft.  Single-Track  Deck  Plate  Girder  Span 

141.  Data.— 

Length  =  90'-0"  c.c.  end  bearings. 

Dead  Load  =  12  x  90  +  550  =  1,630  Ibs.  per  ft.  of  span. 

Live  Load,  Cooper's  E50. 

Specifications,  A.  R.  E.  Ass'n. 

142.  Calculations. — For  the  main  girders  of  this  span  we  have: 

Maximum  End  Shear:  Maximum  Moment: 

D=    36,700  Ibs.  D=    9,902,000  in.  Ibs. 

L=  171,500  Ibs.  L  =  40,057,000  in.  Ibs. 

7=132,000  Ibs.  7=30,813,000  in.  Ibs. 

340,200  Ibs.  80,772,000  in.  Ibs. 

Assuming  the  web  to  be  TV  thick,  we  have 


.,  AKK      80,772,000 

.r  =  1.055A/      '       ' =  113  ins.   (about) 

\  AX  16,000 

for  the  economic  depth.  But,  as  a  few  inches  either  way  from  the  theo- 
retical depth  in  case  of  such  deep  girders  will  not  materially  affect  the 
design,  it  will  be  better  to  take  108"  as  the  depth,  as  this  will  give  us  a 
108"  web,  which  is  a  very  common  plate,  much  more  so  than  the  112" 
or  113".  So  we  will  assume  a  108"  x  Ty  web. 

There  are  three  types  of  flanges  suitable  for  this  girder:  One  made 
up  of  6"  x  6"  angles,  side  plates,  and  cover  plates,  as  shown  in  Fig.  176; 
one  made  up  of  8"  x  8"  angles  and  cover  plates,  as  ordinary  flanges;  or 
one  made  up  of  4 — 6"  x  6"  angles  and  side  plates  for  the  top  flange  and 
8"  x  8"  angles  and  cover  plates  for  the  bottom  flange,  as  shown  in 
Fig.  177. 

The  flanges  made  up  of  8"  x  8"  angles  and  cover  plates  the  author 
thinks  preferable.  However,  the  design  shown  in  Fig.  177  has  some 
desirable  features.  For  instance,  there  are  no  rivet  heads  on  the  top 
flange  to  interfere  with  the  ties,  as  there  are  no  cover  plates  and  the  top 
laterals  are  connected  to  the  bottom  angles  of  the  top  flange. 

The  stress  sheet  and  detail  shop  drawing  and  bills  for  this  span 
can  be  worked  up  in  the  same  manner  as  shown  above  for  the  50-ft.  span. 

All  spans  over  75  ft.  long  should  be  supported  at  one  end  upon 
rollers.  Fig.  178  shows  the  general  details  for  such  a  bearing,  which  is 
for  the  above  90-ft.  span.  Fig.  179  shows  the  general  details  for  the 
corresponding  fixed  support  for  the  same  span.  Figures  180  and  181 
show  the  shop  details  for  the  girder  shoe,  roller  shoe,  roller  nest,  and 
pedestal. 

In  designing  and  drawing  up  such  a  bearing  as  that  shown  in  Fig. 
178,  the  first  thing  is  to  determine  the  space  taken  up  by  the  rollers.  For 
the  above  90-ft.  span  we  have  340,200*  for  the  maximum  reaction.  The 
minimum  size  of  rollers  is  limited  by  the  specifications  to  6"  diameter, 
and  as  plate  girder  bridges  are  comparatively  small  bridges  we  will  use 
the  minimum  size  roller.  The  allowable  pressure  per  linear  inch  on 


Fig.  178 


c 

o  r 

{ 
\ 

V 

* 

i 

] 

*o 

%  

I 

5^ 

M> 


*t  XL* 

>  *o  Q  D  «o 


§  L 

«! 

4 


2-   O" 


Fig.    179 

250 


0 


s 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


251 


these  rollers  is   600  x  6  =  3,600#.      (See  specifications,  also  Art.   84.) 
We  then  have 


340,200 
3,600 


=  94.5 


for  the  linear  inches  of  roller  required.     If  we  use  five  rollers  we  have 

94.5 

'    =  19  ins.  (about) 


for  the  required  net  length  of  each  roller,  and  adding  4^"  to  each  to 
provide  for  the  slots  in  tdie  ped- 
estal we  have  23%"  for  the  total 
length  of  each  roller,  as  shown  in 
Fig,  178.  All  rollers  having  a 
diameter  of  6"  and  over  are 
usually  flattened  on  the  sides  like 
those  shown  in  Figs.  178  and  181, 
and  are  known  as  segmental 
rollers.  These  rollers  should  be 
placed  quite  close  to  each  other 
so  that  their  sides  will  come  to- 
gether whenever  they  are  rolled 
beyond  the  amount  desired.  Thus 
the  rollers  are  prevented  from  fall- 
ing over.  The  distance  between 
the  rollers  (between  the  sides)  is 
usually  about  J",  as  shown  in  Fig. 
181.  After  determining  the  size 
of  the  roller  nest,  a  preliminary 
drawing  of  the  shoes  and  pedestal 
can  be  made  similar  to  that  shown 
in  Fig.  178,  wherein  the  general 
dimensions  are  made  to  conform 
to  those  of  the  roller  nest,  while 
the  correct  size  of  the  pin  passing 
through  the  shoes  and  the  thick- 
ness of  metal  throughout  are  de- 
termined as  the  work  progresses. 
We  begin  by  deciding  upon  the 
general  form  of  the  shoes.  Then 
we  assume  the  size  of  the  pin  passing  through  them.  Next  we  calculate 
the  required  thickness  of  the  bearings  of  the  shoes  for  the  pin  assumed. 
If  this  seems  satisfactory  we  proceed  with  the  work  of  calculating  the 
stress  on  the  pin  to  determine  whether  the  pin  assumed  is  of  correct  size, 
and  if  the  size  is  found  to  be  correct  we  proceed  farther  with  the  design, 
but  if  found  to  be  incorrect  we  start  over  and  make  some  new  assumption, 
and  so  on,  until  the  design  is  worked  out  satisfactorily. 


'                   2-8" 

r       ?-  o" 

la'                2-  4k"                  % 

E-  Roller  Shoes-  RS. 

^  //*  3"SloH-edho/es.  —  . 

P 

(Gost&ee/) 

1  1  i 

...    J 

\ 

j 

A     \ 

tf 

t 

\ 

i 

•$ 

Fi*.   180 


252 


STRUCTURAL  ENGINEERING 


.  8-*.ii5feeip,l>5.tohoK</rnmiqM;l>roiier*          Taking  the  case  shown  in  Fig. 

*N^  *N«.  "^.  -1   fvr~*        1       .  .1  •  i«i// 


178,  let  us  assume  the  pin  to  be 
in  diameter.  The  total  thickness  of 
the  required  bearing  on  each  shoe  is 
then 

340.200 


24,000 


- 


.|]f  Drilled  holes  in  bars. 


•J 

2>      2- Roller  tfest-RM. 


Ur  /•-/• 

^" 

- 

5p 

» 

//-"Corec//>o/es  - 

--:---:::-.-.:--.-;.-.--• 

Ff= 



S 

ill   1 

Z- Pedestal* -RP 

(Cast-  Sfee/.j 

Fig.    181 


The  central  bearing  of  each  shoe 
will  likely  be  subjected  to  a  little 
more  pressure  than  either  of  the  side 
bearings.  So  we  will  make  them  a 
little  thicker  than  the  side  bearings. 
By  making  the  central  bearing  of 
each  shoe  1J"  thick  and  each  side 
bearing  1-|"  we  have  3J"  for  the 
total  thickness  of  bearing  on  each 
shoe,  which  is  about  equal  to  that 
required,  and  as  the  side  bearings 
are  of  minimum  thickness  for  such 
castings  these  thicknesses  are  quite 
satisfactory  and  hence  will  be  used. 
The  maximum  shear  and  cross 
bending  on  the  pin  occur  at  the  side 
bearings.  For  the  maximum  shear 
we  have 

340,200  x  if  =  109,300  Ibs. 
03 


and  for  the  maximum  shearing  unit  stress  on  the  4J"  pin  we  have 
109,300        109,300 


14.18 


-  7,700  Ibs. 


For  the  maximum  cross  bending  on  the  pin  we  have 

109,300  x  If  "=150,000  inch  Ibs.  (about) 

and  for  the  maximum  fiber  stress  due  to  cross  bending  we  have 
.-_  «0,000  x  2J  =2(M) 


Now,  as  the  allowable  shear  on  pins  is  12,000  Ibs.  per  square  inch  and 
the  allowable  fiber  stress  is  24,000  Ibs.  per  square  inch  (see  specifica- 
tions), it  is  seen  from  the  above  that  the  4J"  pin  assumed  is  amply  strong; 
in  fact  it  could  be  reduced  in  size,  but  as  the  saving  in  cost  would  be 
insignificant  we  will  use  the  size  assumed.  It  also  appears  to  be  about 
the  correct  size. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


253 


In  addition  to  having  sufficient  bearing  on  the  pin  the  shoes  should 
be  deep  enough  and  contain  enough  metal  to  resist  the  cross  bending  to 
which  they  are  subjected. 

Considering  the  girder  shoe  (GS)  we  have  a  case  of  cross  bending 
as  indicated  in  Fig.  182.  For  the  maximum  moment  we  have 


340,200 


inch  lbs    (about), 


which  occurs  at  the  vertical  transverse  section  c-c  through  the  pin  hole. 
The  moment  of  inertia  of  this  cross-section  of  the  shoe  (not  subtracting 
the  pin  hole)  is  about  390.  Then  for  the  maximum  compressive  unit 
stress  due  to  cross  bending  we  have 


638,000  x  6| 
390 


=  12,200  lbs. 


and  for  the  maximum  tensile  unit  stress  we  have 

638,000  x  3j 


390 


-5,500  lbs. 


This  shows  that  the  girder  shoe  is  amply  strong,  as  far  as  cross  bending 
is  concerned,  as  16,000  lbs.  unit  stress  is  permissible.  The  bending  on 
the  roller  shoe  (RS)  can  be  determined  in  the  same  manner.  Further 
designing  of  the  shoes  is  very  much  a  matter  of  common  judgment. 

Regarding  the  above  calculations 
it  may  first  appear  that  the  material  cut 
out  of  the  vertical  section  c-c  by  the  pin 
hole  should  be  deducted  from  the  sec- 
tion in  determining  the  moment  of 
inertia.  But  this  is  far  from  being  cor- 
rect, as  the  top  half  of  the  pin  is 
assumed  to  transmit  24,000  lbs.  per 
square  inch  against  the  shoe — this  we 
can  rely  upon — and  owing  to  the  lateral 
deformation  of  the  pin  there  will  be 
some  horizontal  pressure  (perhaps  equal 
to  Poisson's  ratio) ;  so,  taking  all 
in  all,  the  above  assumption  is  fairly 

The  designing  of  the  pedestal  (RP}  is  very  much  a  matter  of  com- 
mon judgment.  The  pedestal  should  be  at  least  4  ins.  or  5  ins.  high  in 
order  to  have  the  rollers  above  snow,  slush,  and  dirt,  and  should  be  broad 
and  long  enough  to  support  the  load  coming  on  them  without  producing 
a  greater  pressure  upon  the  masonry  than  600  lbs.  per  square  inch. 
There  should  be  open  slots  extending  down  from  the  top  as  shown  in 
Fig.  178,  so  that  dirt  will  not  accumulate  between  the  rollers.  The 
longitudinal  circular  holes  should  be  cored  in  the  casting  and  the  slots 
cut  from  the  solid  when  the  pedestals  are  being  finished  in  the  machine 
shop.  This  will  guard  against  the  pedestal  warping  when  cooling.  The 
details  of  the  fixed  bearing  shown  in  Fig.  179  are  considered  to  be  self- 
explanatory. 


UUUiii  UlliUJl! 


254 


STRUCTURAL  ENGINEERING 


143.  Flange  Splices. — The  flanges  of  all  plate  girders  over  90  ft. 
long  must  be  spliced,  as  that  is  about  the  maximum  length  of  angles  and 
cover  plates  obtainable.  These  splices  should  preferably  be  symmetrically 
arranged  in  reference  to  the  center  of  the  span.  The  flange  angle  splices 
should,  as  a  rule,  be  nearer  to  the  ends  of  the  girder  than  to  the  center 
of  the  span,  and  only  one  flange  angle  should  be  cut  off  at  a  splice.  That 
is,  one  flange  angle  should  be  spliced  on  one  side  of  the  girder  on  one 
side  of  the  center  of  the  span  and  the  other  flange  angle  should  be 
spliced  on  the  other  side  of  the  girder  at  a  corresponding  point  on  the 
other  side  of  the  center  of  the  span,  as  shown  in  Fig.  183.  Here  the 
near  flange  angle  is  spliced  at  D,  one  part  extending  from  D  to  B,  and 
the  other  part  from  D  to  A,  while  the  far  flange  angle  is  spliced  at  E, 
one  part  extending  from  E  to  A  and  the  other  part  from  E  to  B. 


Sechm-SS 

Fig.   183 


The  splice  of  a  flange  angle  should  be  made  by  means  of  an  angle 
having  the  same  net  area  of  cross-section,  as  the  flange  angle.  The  splice 
angle  must  be  ground  to  fit  the  fillet  of  the  flange  angle  and  its  legs  cut 
so  as  not  to  project  beyond  the  legs  of  the  flange  angle,  as  indicated  at 
section  S-S.  (Fig.  183.)  It  is  general  practice  to  use  two  splice  angles 
at  each  splice,  one  on  each  side  of  the  girder.  The  splice  angle  on  the 
side  of  the  flange  angle  spliced  should  be  of  sufficient  cross-section  in 
every  case  to  splice  that  angle.  The  splice  angle  on  the  other  side  of 
the  girder  is  used  simply  to  balance  the  flange  section  at  that  point. 

The  cover  plates  are  usually  spliced  by  extending  the  adjacent  cover 
plates  a  short  distance  beyond  their  theoretical  length,  as  shown  in  Fig. 
183.  As  a  rule,  the  cover  plates  next  to  the  flange  angles  are  the  only 
ones  that  need  to  be  spliced.  Cover  plates,  of  course,  can  be  spliced  by 
simply  using  a  short  plate  having  the  same  area  of  cross-section  as  the 
plate  spliced,  but  usually  this  presents  an  unsightly  appearance. 

As  an  example  in  figuring  splices,  let  the  two  flange  angles  shown  in 
Fig.  183  be  2 — l_s  6"  x  6"  x  f  ",  as  indicated.  For  the  net  area  of  either 
of  these  flange  angles  we  have  (using  J"  rivets)  7.11  - 1.25  =  5.86n//.  Then 
for  the  allowable  stress  in  each  flange  angle  we  have  5.86  x  16,000  = 
93,700*.  Then  for  the  number  of  J"  shop  rivets  required  on  each  side 
of  the  splice  (in  the  splice  angle)  we  have  93,700 -j- 7,200  =  13  rivets  (in 
single  shear).  As  is  seen,  14  are  used — 7  in  the  vertical  leg  and  7  in  the 
horizontal  leg.  The  splice  angle  at  each  splice  should  have  5.86n"  net. 
Using  a  6"x6"x{J"  angle,  and  deducting  1.37n"  for  rivet  holes  and 
0.86n"  for  the  cutting  of  the  legs  and  for  the  grinding  to  fit  fillet,  we 
have  7.78  -1.37-  0.86  =  5.55n"  net,  which  is  about  correct. 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES 


255 


The  calculation  for  the  splicing  of  a  cover  plate  is  simply  a  matter 
of  determining  the  number  of  rivets  required  in  single  shear  to  develop 
the  strength  of  the  plate  spliced.  The  student  should  have  no  trouble 
in  doing  that. 

144.  Graphical  Determination  of  Live-Load  Shear  and  Bending 
Moment  on  Deck  Plate  Girder  Bridges. —  The  analytical  method  out- 
lined in  Art.  133  is  generally  used,  as  it  is  simple  in  application  and  the 
results  obtained  are  absolutely  numerically  accurate,  yet  the  graphical 
method  is  fully  as  simple  in  application  and  the  results  obtained  are 
quite  accurate  enough.  The  graphical  method  at  least  affords  an  excel- 
lent means  of  checking  results  obtained  analytically,  and  if  for  no  other 
reason  than  this  the  engineer  should  be  familiar  with  the  method.  In 
fact,  there  are  two  methods:  that  of  influence  lines  and  the  one  wherein 
the  equilibrium  polygon  is  used. 


so'-o' 


Fig.   184 

The  application  of  influence  lines  is  fully  given  in  Arts.  100  and 
101.  As  an  application  of  the  equilibrium  polygon  let  us  take  the  case 
of  the  50-ft.  span  treated  analytically  in  Art.  133,  where  Cooper's  £50 
loading  is  used.  Let  AB  (Fig.  184)  represent  the  span  drawn,  say,  to 
J"  scale.  From  inspection  of  the  diagram  in  Table  A  (in  the  back  of 
this  book)  we  can  see  that  the  maximum  end  shear  will  occur  when 
wheel  2  is  at  the  end  of  the  girder  and  wheels  2  to  10  (inclusive)  are 
on  the  span,  as  shown  in  Fig.  184.  After  the  loads  are  thus  placed,  the 
next  thing  to  do  is  to  lay  off  the  load  line  uv  (say,  to  a  -gSr-in.  scale)  at 
(6),  and  construct  the  ray  diagram,  as  explained  in  Art.  95,  and  draw 
the  corresponding  equilibrium  polygon  ab...na  at  (a).  Then  by  draw- 
ing from  0  (at  6)  the  line  OE  parallel  to  the  closing  line  an,  we  have 
the  maximum  end  shear  given  by  the  part  of  the  load  line  extending  from 
E  to  u  which  can  be  scaled,  using  the  same  unit  of  scale  as  was  used  in 
laying  off  the  load  line. 

To  determine  the  shear  at  any  point  K  of  the  girder,  place  the  point 
K  under  wheel  2  by  moving  the  girder,  so  to  speak,  to  the  left,  so  that 


256 


STRUCTURAL  ENGINEERING 


the  end  A  will  be  at  A'  and  the  other  end  will  be  at  Bf  and  wheels  1  to  7 
will  be  on  the  span.  Then  add  wheel  1  to  the  load  line  and  draw  the 
ray  wO  and  extend  the  equilibrium  polygon  on  to  s  and  we  have  the 
equilibrium  polygon  sta . . .  fs's.  Then  draw  OE'  in  the  ray  diagram 
parallel  to  the  closing  line  **'  and  we  have  the  maximum  shear  at  point 
K  given  by  the  part  of  the  load  line  extending  from  E'  to  u.  In  this 
manner  the  maximum  shear  at  any  point  on  the  span  can  be  determined. 
To  determine  the  maximum  bending  moment  place  the  loads  on  the 
span,  as  shown  in  Fig.  185,  so  as  to  satisfy  the  criterion  for  maximum 
bending  moment  as  per  Art.  88.  Then  construct  the  ray  diagram  as 


Fig.   185 

shown  at  (6)  and  draw  the  corresponding  equilibrium  polygon  ab  . . .  ma, 
and  we  obtain  the  maximum  moment  by  multiplying  the  ordinate  e'e 
under  wheel  13  (e'e  must  be  scaled  off  in  feet  or  inches  to  the  same 
scale  as  was  used  in  laying  off  the  length  of  span  at  (a)  and  spacing 
the  loads)  by  H,  the  pole  distance  which  is  laid  off  in  pounds  to  the 
same  scale  as  the  load  line.  If  the  ordinate  e'e  is  taken  in  feet  the 
bending  moment  will  be  in  foot  pounds,  and  if  taken  in  inches  the  bending 
moment  will  be  in  inch  pounds. 

Such  a  diagram  as  shown  in  Fig.  186  is  very  convenient  if  the  work 
to  be  done  is  extensive  enough  to  warrant  the  drawing  of  it.  In  offices 
where  one  loading  is  used  for  all  bridges  it  is  advisable  to  draw  up  such 
a  diagram  of  the  loading  on  a  sheet  of  thick  cardboard,  inking  in  all 
lines  shown  in  Fig.  186  except  the  closing  lines.  The  closing  lines  can 
be  drawn  lightly  in  pencil  as  needed  and  then  erased.  In  this  way  the 
diagram  can  be  used  in  determining  the  shears  and  moments  for  any 
number  of  bridges.  The  diagram  shown  in  Fig.  186  is  for  Cooper's  J£40 
loading.  The  loads  are  spaced  off  to  -^-in.  scale  on  the  horizontal  line 
at  the  top  of  the  sheet.  Then  the  load  line  LL  is  laid  off  to  a  Tsfo-in.  scale 
and  the  ray  diagram  is  constructed,  as  shown,  and  the  corresponding 
equilibrium  polygon  ABC  is  drawn,  and  then  the  scale  lines  are  drawn 
below  this,  as  shown. 

To  show  the  manner  of  using  the  diagram  in  Fig.  186,  let  us  take 
the  case  of  a  50-ft.  span.  The  maximum  moment  due  to  Cooper's  loading 
in  the  case  of  most  deck  plate  girder  bridges  will  occur  under  one  of  the 
four  wheels  11  to  14  (this  we  obtain  from  experience).  So,  beginning  with 
zero  at  wheel  13  we  lay  off  the  scale  line  HII  into  f)-ft.  units.  To  determine 
the  maximum  bending  moment  in  this  50-ft.  span  let  us  start  by  assuming 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


257 


that  the  span  extended  out. 25  ft.  each  side  of  wheel  13.  Then  drawing 
the  closing  line  1-1  we  have  the  equilibrium  polygon  l-$-l,  and  bv 
scaling  off  the  maximum  ordinate  between  the  closing  line  1-1  and  the 
curve  1-5-1  and  multiplying  it  by  the  pole  distance  given  in  the  ray 
diagram  we  obtain  the  maximum  bending  moment  for  the  span  in  that 
position.  By  moving  the  span  5  ft.  to  the  right  and  drawing  the  closing 
line  2-2  we  have  the  equilibrium  polygon  2-B-2  and  by  scaling  off  the 
maximum  ordinate  and  multiplying  it  by  the  pole  distance  we  obtain 
the  maximum  bending  moment  for  the  span  in  that  position,  and  moving 
the  span  to  the  left  the  equilibrium  polygon  3-B-3  is  drawn  and  the 
maximum  bending  moment  for  the  span  in  that  position  can  be  deter- 
mined. Now,  by  simply  drawing  a  few  of  these  sub-equilibrium  polygons 
the  absolute  maximum  ordinate  can  be  ascertained  from  the  intersection 
of  the  closing  lines  with  the  verticals  through  the  wheels.  The  maximum 


35'    30'    25'    20'   i/5'   Jo'    \S'    \O       \f'    10'     IS'   ZO'  2S' 

!         i         '         !  I 


ordinate  will  always  be  under  a  wheel,  and  after  a  sufficient  number 
of  closing  lines  are  drawn  the  maximum  ordinate  can  be  quickly  deter- 
mined by  the  aid  of  a  pair  of  dividers  and  then  scaled  off  and  multiplied 
by  the  pole  distance.  In  this  way  the  maximum  bending  moments  are 
determined  without  bothering  with  the  criterion  for  maximum  moment, 
as  is  necessary  when  other  methods  are  used. 

To  determine  the  maximum  end  shear  on  the  50-ft.  span,  place 
wheel  2  at  the  end  of  the  span.  Now  using  the  scale  line  SS,  the  span 
will  extend  50  ft.  to  the  right  of  wheel  2  and  we  obtain  an  equilibrium 
polygon  which  has  the  closing  line  nn'.  Then  drawing  Pr  in  the  ray 
diagram  parallel  to  this  closing  line  we  have  the  maximum  end  shear 
or  reaction  given  by  the  part  of  the  load  line  extending  from  r  to  e.  By 


258 


DESIGN   OF  SIMPLE   RAILROAD   BRIDGES  259 

the  following  construction  the  end  shears  can  be  obtained  more  quickly 
than  if  scaled  from  the  ray  diagram:  From  n  (on  the  equilibrium 
polygon)  lay  off  the  horizontal  line  nm  equal  to  the  pole  distance  EP 
and  draw  the  vertical  line  kk'  through  m  and  prolong  the  segment  gn, 
of  the  equilibrium  polygon,  until  it  intersects  this  vertical  line  at  k' '. 
Then  we  have  the  triangle  nmkf  equal  to  the  triangle  Pre  (in  the  ray 
diagram),  and  hence  xk'  is  equal  (by  scale)  to  the  maximum  end  shear 
on  the  50-ft.  span.  Likewise,  xf  h'  and  x"kf  are  equal,  respectively,  to 
the  maximum  end  shear  on  a  60-ft.  and  75-ft.  span. 

The  equilibrium  polygon  ABC  can  be  used  in  general  to  determine 
the  shear  at  any  point  in  a  span  in  the  manner  explained  in  the  case 
shown  in  Fig.  184. 

Problem  1.  Construct  a  diagram  for  Cooper's  E50  loading  similar 
to  the  diagram  shown  in  Fig.  186  and  determine  from  it  the  maximum 
end  shears  and  bending  moments  on  a  40-ft.,  50-ft.,  60-ft.,  70-ft.,  80-ft., 
and  90-ft.  span. 

DRAWING  ROOM  EXERCISE  NO.  3 

Design  a  60-ft.  single-track  deck  plate  girder  railroad  bridge  and 
make  a  stress  sheet  for  the  same.  The  finished  drawing,  similar  to  that 
shown  in  Fig.  169,  is  to  be  upon  an  18"  x  24"  sheet  of  tracing  cloth. 

Data: 

Length  =  60 '-0"  c.c.  end  bearings. 

Width  =  6'-6"c.c.  girders. 

Height,  to  be  determined  by  student. 

Dead  Load,  to  be  determined  by  the  student. 

Live  Load,  Cooper's  E5Q. 

Specifications,  A.  R.  E.  Ass'n. 

DRAWING  ROOM  EXERCISE  NO.  4 

Make  a  shop  drawing  and  shop  bills  for  the  60-ft.  bridge  specified 
in  Drawing  Room  Exercise  No.  3.  The  finished  drawing,  similar  to  that 
shown  in  Fig.  171,  is  to  be  upon  a  24"  x  36"  sheet  of  tracing  cloth. 

THROUGH  PLATE  GIRDER  BRIDGES 

145.  Preliminary. —  Through  plate  girder  bridges  are  used,  as  a 
rule,  only  when  the  under  clearance  will  not  permit  of  the  use  of  a  deck 
span — owing  to  the  cost  of  the  through  bridge  being  considerably  more 
than  the  deck  bridge.  The  ordinary  through  plate  girder  bridge  is  com- 
posed of  two  main  girders  and  a  floor  system  composed  of  longitudinal 
beams  (or  girders),  which  support  the  ties  and  are  known  as  stringers, 
and  cross  beams,  which  support  the  stringers  and  are  known  as  floor 
beams. 

An  isometric  view  of  an  ordinary  single-track  through  plate  girder 
span  is  shown  in  Fig.  187,  where  the  names  of  the  different  parts  of 
the  structure  are  given. 

Double-track  through  plate  girder  bridges  are  usually  the  same  in 
construction  as  the  single-track  bridges,  except  the  main  girders  in  the 
double-track  structures  are  farther  apart  and  the  floor  system  provides 
for  the  two  tracks,  which  requires  twice  as  many  stringers. 


260  STRUCTURAL  ENGINEERING 

Complete  Design  of  a  60-Ft.  Single-Track  Through  Plate  Girder  Span 

146.  Data.- 

Length  =  4  panels  @  15'-0"  =  60'-0"  c.c.  end  bearings. 
Width  =  IS'-G"  c.c.  main  girders. 
Assumed  Dead  Load: 

For  main  girders,  (13  x  60  +  600  4  400)  =  1,780  Ibs.  per  ft. 

of  span. 
For  stringers,  (12  x  15  +  100  +  400)  =  680  Ibs.,  say  700  Ibs., 

per  ft.  of  span. 

Live  Load,  Cooper's  £50  loading. 
Specifications,  A.  R.  E.  Ass'n. 

147.  Design  of  15-Ft.  Stringers.  —  It  is  usually  necessary  to  make 
the  stringers  in  through  plate  girders  just  as  shallow  as  good  details  will 
permit,  owing  to  the  under  clearance  being  limited.     This  can  be  accom- 
plished best  by  the  use  of  I-beams,  two  or  more  under  each  rail.     So  we 
will  use  I-beams. 

For  dead-load  moment,  using  the  load  assumed  in  Art.  146,  we  have 


M=    -xx!5x  12  =  118,000   inch  Ibs. 
o        £ 

For  live-load  moment  we  have 

M'  =  1,875,000  inch  Ibs.  (same  as  Art.  130). 
For  impact  we  have 

7=1,785,000  inch  Ibs.    (same  as  Art.  130). 
Then  we  have  for  the  total  bending  moment 

118,000  +  1,875,000  +  1,785,000  =  3,778,000  inch  Ibs. 
Then  for  the  section  modulus  we  have 
3,778,000 


16,000 


=  236.     (See  Art.  105.) 


This  calls  for  2 — Is,  20"  x  70*,  under  each  rail.     That  is,  each  stringer 
is  to  be  composed  of  2—20"  x  70#  I-beams. 

For  dead-load  end  shear  on  each  stringer  we  have 

R  =^T  x  ¥  =2,625  Ibs.,  say  2,600  Ibs. 
£        <i 

For  live-load  end  shear  we  have 

fl'  =  50,000  Ibs.  (same  as  Art.  130) 
and  for  impact  we  have 

I  =  47,600  Ibs. 
Then,  for  the  total  end  shear  we  have 

2,600  +  50,000  +  47,600  =  100,200  Ibs. 


DESIGN  OF  SIMPLE  KAILEOAD  BRIDGES 


261 


Preliminary  estimate  of  weight  of  stringers. 

4—  Is    20"x70#xl5'-0"  =  4,200  Ibs. 

1_[    9"x20#x3'-9"  =       75  Ibs. 

2  —  end  con.  Ls  on  9"  [s  =       21  Ibs. 

2—  [s  '  12"  x  25#  x  l'-3"  =      62  Ibs. 

Total  weight  of  one  panel  =  4,358  Ibs. 

Weight  of  stringers  per  ft.  of  span  =  4,358-=-  15  =  289  Ibs.  metal. 
Weight  of  deck  =400  Ibs.  metal 

Total  dead  load  for  stringers  per  ft.  of  span  =  689  Ibs.  metal. 

This  shows  that  the  700  Ibs.  dead  load  assumed  above  is  about 
correct. 

148.  Design  of  Intermediate  Floor  Beams.  —  The  dead  load  on 
these  beams  consists  of  the  dead  weight  applied  to  them  by  the  stringers 
and  also  the  weight  of  the  floor  beams  themselves. 

The  dead  load  from  the  stringers  is  concentrated  on  the  floor  beams 
at  the  points  where  the  stringers  are  connected  to  them.  This  concen- 
tration at  each  point  is  equal  to  twice  the  dead-load  end  shear  on  one 
stringer  which  is  given  above  as  2,600  Ibs.  So  each  concentration  is 
2,600x2  =  5,200  Ibs.  The  weight  of  the  floor  beam  is  a  uniform  load. 
Such  floor  beams  usually  weigh  about  3,600  Ibs.  each.  So  we  will  assume 
that  weight.  Then  the  dead  load  on  any  intermediate  floor  beam  will  be 
as  shown  in  Fig.  187A,  where  A-B  represents  the  beam. 

It  is  readily  seen  from  Fig.  187  A  that 
zero  shear  occurs  at  the  center  of  the  beam 
and   hence   that   is   the   point   of   maximum 
moment.      As   the   bending   moment   on  the 
floor  beam  due  to  the  two  concentrated  loads 
is  constant  between  the  stringers  it  is  cus- 
tomary to  multiply  one  concentration  by  its 
distance  from  the  end  of  the  beam  to  obtain 
the  bending  moment  due  to  these  loads. 
To  show  the  correctness  of  this  method  let  AB,  Fig.  188,  represent 
a  floor  beam  supporting  the  two  equal  concentrated  loads  P,  as  shown. 
Taking  moments  about  the  center  of  the  beam  we  have,  since  R  =  P, 


1 

I 

W/////////////A 

''//360'GW/A 

W//////////M 

R    5-3" 

C5'-0" 

,s'-e» 

—  n 
Fig.   187A 

That  the  moment  between  the  loads  is  constant  can  also  be  shown  very 

readily  by  graphics  :    Lay  off  the  load  line  VWS 

(Fig.  188)  and  construct  the  ray  diagram  VOS 

and  draw  the  corresponding  equilibrium  polygon 

efghe.     Now,  as  the  reactions  are  equal  for  the 

two  loads,  the  closing  line  eh  must  be  parallel  to 

the  ray  OW,  and  as  the  segment  fg  is  parallel 

to  this  ray  also,  it  is  seen  that  the  ordinates  of 

the   equilibrium  polygon   are  constant   between 

the  two  loads,  and  hence  the  moment  between 

them  must  be  constant.     Now  going  back  to  the  rigr.  188 


262  STRUCTURAL  ENGINEERING 

problem  in  hand,  we  have  the  moment 

M  =  5,200  x  63  =  327,600  inch  Ibs. 
from  the  concentrated  load  and 
1      3,600 


M/  =zr x  ^nrr  x  15-5  x  13  =  83>700  inch  lbs- 

8        15.5 

from  the  weight  of  the  floor  beam,  making  a  total  maximum  of  411,300 
inch-lbs.  dead-load  bending  moment. 

It  is  obvious  that  the  maximum  live-load  bending  moment  on  the 
floor  beam  will  occur  when  the  maximum  live-load  concentrations  from 
the  stringers  on  to  the  floor  beam  occur.  So  it  is  first  necessary  to  deter- 
mine the  position  of  the  wheels  when  the  maximum  live-load  concentra- 
tions occur. 

Let  Fig.  189  represent  two  adjacent  panels  of  stringers  with  a  floor 
•  beam  at  each  of  the  ends,  A,  B, 

*  |*_ z        ..       and    C.     Let   I   be   the   length   of 

jJirX  s~\  \  A"^  panel    AB    and    r    the    length    of 

panel   BC,   as   indicated.     Let    W 


o  o 


Y* My  be  the  total  weight  of  the  wheels 


Fi     189  in  panel  AB,  the  center  of  gravity 

of  which  we  will  assume  is  x  dis- 

tance from  A,  and  let  W  be  the  total  weight  of  the  wheels  in  panel  BC, 
the  center  of  gravity  of  which  we  will  assume  is  s  distance  from  C. 
Let  r  be  the  concentration  on  the  floor  beam  at  B  due  to  the  wheels  in 
the  panel  AB,  and  let  rf  be  the  concentration  due  to  the  wheels  in  the 
panel  BC,  and  let  R  be  the  concentration  on  the  floor  beam  at  B  from 
the  wheels  in  both  panels.  Then  we  have 

R  =  r+r'  ...........................................  (1). 

Taking  moments  about  A  we  have 

_Wx 

I 

and  taking  moments  about  C  we  have 

W  <? 


V 
Then  substituting  these  values  in  (1)  we  have  the  general  expression 


for  the  concentration  on  the  floor  beam  at  B.  Now,  if  the  wheels  roll 
to  the  left,  r  will  decrease  and  r'  increase  provided  no  wheels  pass  B, 
and  just  the  reverse  is  true  if  the  wheels  roll  to  the  right.  Now  if  the 
weight  of  the  wheels  in  the  two  panels  is  such  that  the  increment  of  r 
is  just  equal  to  the  increment  of  r'  for  a  very  slight  movement,  it  is  evi- 
dent that  the  increment  of  the  concentration  R  will  be  zero  and  hence  R 
will  then  be  a  maximum,  for  otherwise  it  would  be  either  increasing  or 
decreasing.  If  increasing,  it  evidently  has  not  reached  the  maximum; 
and  if  decreasing,  it  has  passed  the  maximum.  So,  differentiating  equa- 
tion (2)  we  have 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  263 


/  /' 

But  dz  —  dx.    Then  we  have 

dR      W      W  W     W 

-37-  =  -7-  +  -77-  =  0     or   -y-  =  ~YT  - 
dx       I         I  I         I 

That  is,  the  unit  load  in  one  panel  will  be  equal  to  the  unit  load  in  the 
other  panel  when  the  maximum  concentration  on  the  floor  beam  occurs. 
If  the  panels  are  of  equal  length,  as  they  are  here,  we  have 

w=w. 

That  is,  the  maximum  concentration  on  the  floor  beam  will  occur  when 
the  load  in  one  panel  is  equal  to  the  load  in  the  other.  Now  the  increment 
of  R  can  change  sign  only  when  a  wheel  passes  the  floor  beam  at  B. 
So  there  will  be  a  load  at  that  point  when  the  maximum  concentration 
occurs.  It  is  evident  that  this  criterion  for  maximum  concentration  on  the 
floor  beam,  so  far  established,  can  be  satisfied  by  most  any  group  of 
wheels,  but  it  is  also  evident  that  the  absolute  maximum  concentration  will 
occur  when  the  heaviest  group  of  wheels  is  near  the  floor  beam.  So  the 
whole  criterion  for  maximum  concentration  on  an  intermediate  floor  beam 
can  be  stated  as  follows: 

The  maximum  concentration  will  occur  when  the  heaviest  group  of 
wheels  is  near  the  floor  beam  and  one  wheel  at  it,  and  when  the  load  in 
one  panel  is  equal  to  the  load  in  the  other. 

The  next  thing  in  our  case  is  to  satisfy  this  criterion.  Referring  to 
Table  A  we  can  see  that  wheels  1  to  5  are  as  heavy  a  group  of  wheels  as 
can  be  placed  on  two  adjacent  panels. 

Let  us  place  them  as  shown  in  Fig.  190.  Then  the  load  in  panel 
AE-  30,000  Ibs.,  and  that  in 
BC  =  40,000  Ibs.  This  does  not 
seem  to  be  very  close  to  the  re- 
quirement— that  the  loads  in  the 
two  panels  be  equal.  But  by  trial 
it  will  be  found  to  be  as  close  as 
possible  and  still  satisfy  the  first 
part  of  the  criterion  at  the  same  Fig.  100 

time.     It  is  not  often  that  this 

criterion  can  be  satisfied  absolutely  as  to  the  loads  in  the  panels  being 
equal.  We  should,  however,  place  the  loads  so  that  the  criterion  is  as 
nearly  satisfied  as  possible.  In  testing  for  the  criterion  the  load  at  the 
floor  beam  can  be  considered  as  being  equally  divided  between  the  two 
panels;  that  is,  one-half  of  its  weight  being  considered  in  each  panel, 
or  it  can  be  ignored  as  was  done  above.  It  will  make  no  difference  which 
way  it  is  considered. 

So,  taking  the  position  shown  in  Fig.  190  as  satisfying  the  criterion  and 
taking  moments  about  C,  we  have 

20,000  pTF^I  =20,000  Ibs., 


264 


STRUCTURAL  ENGINEERING 


and  taking  moments  about  A  we  have 

10,000  x  2  +  20,000  x  10 
15 

Then  adding  these  two  concentrations  to  the  20,000-lb.  load  at  the  floor 
beam,  we  have 

R  =  20,000  +  14,660  +  20,000  =  54,600  Ibs. 

for  the  maximum  floor  beam  concentration  due  to  Cooper's  £40  loading, 
and  multiplying  this  by  50/40  to  reduce  it  to  £50,  we  have 

^x  54,600  =  68,200  Ibs.  (about) 
for  the  maximum  concentration. 

§|  §.  Then  the  maximum  live  load  on  the  floor 

a,  beam  will  be  as  shown  in  Fig.  191. 

°  For   the    maximum    live-load    bending    mo- 

,  we  have 

M  =  68,200  x  63  =  4,297,000  inch  Ibs., 
and  for  the  impact  we  have 

x  4,297,000  =  3,906,000  inch  Ibs. 
oU  +  oUU 

Now  adding  the  above  maximum  dead-  and  live-load  bending  moments 
and  impact  together  we  have 

411,000  +  4,297,000+3,906,000  =  8,614,000  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  floor  beam  for  which  the 
beam  must  be  designed  to  resist. 

For  the  maximum  end  shear  on  the  floor  beam  due  to  dead  load  we 
have 


T7K 

«U      £~"O        L 

|r?      rt 

IS'-  6" 

| 

~\ 

Fig.   191 

a 

7 

300 

5,200 


=  7,000  Ibs.   (See  Fig.  187A.) 


From  live  load  we  have  68,200  Ibs.,  as  shown  above  (Fig.  191),  and  from 
impact  we  have 

68,200  x 


Then  adding  the  above  dead-  and  live-load  end 
shears  and  impact  together,  we  have 

7,000  +  68,200  +  62,000  =  137,200  Ibs. 

for  the  total  maximum  end  shear. 

The  next  thing  is  to  design  the  floor  beam. 
The  flanges  of  such  floor  beams  are  usually  com- 
posed  of  6"  x  6"  angles,  and  the  stringers  should 
fit  in  between  the  flanges  as  shown  in  Fig.  192  so 
as  to  avoid  fillers.  So  giving  a  J"  clearance  at  both  the  top  and  bottom 
of  the  stringers,  as  shown,  we  obtain  »  floor  beam  22%"  deep.  The 


Fig.  102 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  265 

distance  from  the  back  of  the  6"  x  6"  flange  angles  to  their  center  of 
gravity  could  be  obtained  from  a  handbook  or  from  Table  6  provided 
the  exact  weight  of  them  were  known.  However,  we  can  obtain  the 
average  from  this  source.  So  let  us  take  1.78"  as  the  average.  Then 
for  the  approximate  effective  depth  we  have 

32.5 -1.78x2  =  28.94  ins. 
Now  dividing  this  into  the  maximum  bending  moment  given  above  we  have 

8ii4;°00  =  898,000  Ibs.  (about) 

/CO. 94: 

for  the  flange  stress.    Then  we  have 

298,000 


16,000 


=  18.6  sq.  ins. 


for  the  net  flange  area  required  minus  one-eighth  of  the  area  of  the  cross- 
section  of  the  web.  Let  us  assume  the  web  to  be  32"  xj".  Then  one- 
eighth  of  the  area  of  web  =  2n". 

Then  for  the  make-up  of  each  flange  we  have 

Gross  Net 

2— Ls  6"  x  6'  x  ff"  =  18.18n"-1.62  =  16.56n" 
area  of  web  =    2.00n" 


18.56n" 

Now   for  the  actual  effective  depth   (since  we  know  the  weight  of  the 
flange  angles)  we  have 

32.5-1.8x2  =  28.9  ins., 

which  is  practically  the  same  as  assumed  above. 

As  the  floor  beam  has  no  regular  stiffeners  (the  stringers,  however, 
stiffen  it)  the  vertical  distance  between  the  flange  angles  can  be  taken  as 
d  in  Formula  (1),  Art.  118.  Then  we  have 

20.5  =i  (12,000-*), 

from  which  we  obtain 

5  =  10,360  Ibs.,  say,  10,000  Ibs. 

Now  dividing  this  into  the  maximum  end  shear  given  above  we  have 
137,200 


10,000 


=  13.72  sq.  in. 


for  the  required  area  of  the  cross-section  of  the  web.  The  web  assumed 
above  has  16n",  which  is  a  little  more  than  required,  but  if  a  thinner  web 
be  used  the  flange  rivets  would  be  quite  close  together,  perhaps  too  close. 
This  trouble  often  occurs  in  such  shallow  floor  beams.  So  we  will  use  the 
l" 


266  STRUCTURAL  ENGINEERING 

Preliminary  estimate  of  the  weight  of  an  intermediate  floor  beam.* 

1—  web  32"  x  y  x  15.5'  x  54.4#  .........................  =    843  Ibs. 

4—  Ls  6"  x  6"  x  if"  x  15.5'  x  31*  .......................  =1,922  Ibs. 

Gusset  plates  (1  plate  60"  x  \"  x  4'-0"  for  the  two)  @  102# 

per  ft  ............................................  =  408  Ibs. 

4—  Ls  3y  x  3J"  x  |"  x  4'-0"  x  8.5*   (on  gusset  plates  or 

brackets)   ........................................  =  136  Ibs. 

2—  Ls  31"  x  W  x  f"  x  2'-6"  x  8.5#  (end  con.  angles)  ......  =  42  Ibs. 

4—  special  plates  16"  x  y  x  l'-8"  x  27.20*  ...............  =  181  Ibs. 

3,532  Ibs. 
2.5%  for  rivets  ..........................................       88  Ibs. 

Total    .........................................  3,620  Ibs. 

This  shows  that  the  3,600  Ibs.  assumed  weight  of  the  floor  beam  is 
very  close  to  the  actual  weight  and  hence  no  recalculations  are  necessary 
as  10  per  cent  variation  either  way  is  permissible. 

149.  Design  of  End  Floor  Beam.  —  The  designing  of  end  floor 
beams  is  very  much  the  same  as  that  of  intermediate  floor  beams.  The 
concentrations,  however,  are  different  and  the  weight  of  the  beams  them- 
selves is  a  little  less. 

Each  concentration  from  dead  load  is  equal 
to  the  dead-load  end  shear  on  a  stringer  and 
each  live-load  concentration  is  equal  to  the  maxi- 
mum live-load  end  shear  on  the  same.  Then 
assuming  the  weight  of  an  end  floor  beam  to  be 
3,000  Ibs.  we  have  the  loading  on  the  beam  as 
shown  in  Fig.  193,  using  the  end  shears  given 
in  Art.  146.  Then  we  have  for  the  dead-load 
moment. 

M  =  2,600  x  63  =  163,800  inch  Ibs. 
from  the  concentrated  dead  load  and 

M'  =  ^x  3,000x15.5x12  =  69,750  inch  Ibs. 

from  the  weight  of  the  floor  beam,  making  a  total  of  233,550  inch  Ibs.  for 
the  total  maximum  dead-load  bending  moment. 
For  the  live-load  bending  moment  we  have 

M"  =  50,000x63  =  3,150,000  inch  Ibs., 
and  for  the  impact  we  have 


7=3,150,000    1Knn    =  3,000,000  inch  Ibs. 

+ 


Now  adding  these  moments  and  impact  together,  we  have 

233,550  +  3,150,000  +  3,000,000  =  6,383,550  inch  Ibs. 

for  the  total  maximum  bending  moment. 

Assuming  an  effective  depth  of  29"  we  have 

6,383,550 


*  These  preliminary  estimates  are  made  before  the  detail  drawings  are  made  and 
are  intended  to  be  only  approximately  correct. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  267 


for  the  flange  stress.     Then  we  have 
220,000 


16,000 


=  13.75  sq.  ins. 


for  the  net  area  required  for  each  flange  minus  one-eighth  of  the  area  of 
the  cross-section  of  the  web. 

Assuming  the  web  to  be  32"  x  f  "  =  12°",  one-eighth  of  the  area  of 
the  web  is  1.5n//.     Then  for  each  flange  we  have 

2—  Ls  6"  x  6"  x  TV  =  12.86n"  -  1.12  =  11.74n"net 
£  area  of  web  =  1.50n"net 

13.24°"  net 

As  is  seen,  the  net  area  of  this  flange  is  about  0.51n"  less  than  required, 
but  the  6"  x  6"  x  f  "  angles  give  a  flange  about  0.72n"  too  large,  so  the 
above  will  be  used. 

For  the  maximum  end  shear,  as  is  readily  seen  from  Fig.  193,  we  have 

D  =  1,500  +  2,600      =    4,100  Ibs. 
L  =  50,000  Ibs. 


7  -  50,000          r        =  47,600  Ibs. 


Total  101,700  Ibs. 

For  the  allowable  stress  on  the  web  we  have 

20.5=1(18,000-.), 

from  which  we  obtain 

s  =  9,815  Ibs.,  say,  10,000  Ibs. 

Then  dividing  this  into  the  above  shear  we  have 
101,700 


10,000 


=  10.17  sq.  ins. 


for  the  required  area  of  the  web.     So  the  assumed  web  32"  x  §"  =  12n" 
is  satisfactory,  as  the  specifications  permit  of  no  thinner  web. 

The  flange  angles  and  web  as  specified  above  for  the  end  floor  beam 
weigh  about  2,000  Ibs.,  which  is  about  765  Ibs.  less  than  the  flange  angles 
and  web  in  the  intermediate  floor  beam.  Then  as  the  other  parts  are  about 
the  same  as  for  an  intermediate  floor  beam  we  have  3,621  -  765  =  2,856 
Ibs.  for  the  weight  of  the  end  floor  beam.  So  the  assumed  3,000  Ibs.  for 
the  weight  of  an  end  floor  beam  is  within  the  10  per  cent  limit. 

150.  Design  of  the  Main  Girders. — In  the  case  of  through  plate 
girder  bridges  the  live  load  is  applied  to  the  main  girders  only  at  the 
panel  points,  that  is,  where  the  floor  beams  connect  to  the  main  girders, 
just  the  same  as  in  the  case  of  through  truss  bridges.  The  dead  load  is 
applied  to  the  main  girders  in  the  same  way  except  for  the  weight  of  the 
girders  themselves  which  is  uniformly  distributed  along  the  girders.  But 


268 


STRUCTURAL  ENGINEERING 


it  is  usual  practice  to  consider  the  dead  load  as  wholly  applied  at  the 
panel  points  as  about  the  same  result  is  obtained  as  if  the  more  exact 
conditions  were  considered. 

Taking  the  assumed  dead  load  given  in  Art.  146  we  have 

!5  =  13,350  Ibs. 


/v 

for  the  panel  load  of  dead  load  per  girder.  Then  the  dead  load  per  girder 
will  be  as  shown  in  Fig.  194,  where  AB  represents  the  girder.  Now  it  is 
obvious  that  the  maximum  bending  moment  due  to  this  dead  load  will 
occur  under  the  load  at  C.  So  taking  moments  at  C  we  have 

M=  (20,025x30  -13,350x15)  12  =  4,806,000   inch   Ibs. 

for  the  maximum  bending  moment  due  to  dead  load. 

The  live  load  will  be  ap-  * 

plied  at  the  panel  points,  but 
of  course  the  panel  loads  will 
not  be  equal  to  each  other  as 
in  the  case  of  dead  load.  The 
maximum  bending  moment 
will  occur  at  a  panel  point  as 
in  the  case  of  a  truss.  It  is 
readily  seen  that  the  maximum  moment  will  occur  at  the  panel  point  C, 
at  the  center  of  the  span.  Then  the  first  thing  to  do  is  to  place  the 
wheel  loads  so  as  to  obtain  the  maximum  moment  at  that  point  which  is 
simply  the  satisfying  of  the  criterion  for  maximum  moment  as  given  in 
Art.  91.  That  5s,  the  maximum  live-load  bending  moment  on  the  girder 
(which  will  occur  at  C)  will  occur  when  the  average  load  to  the  left  of  C 
is  equal  to  the  average  load  on  the  span. 

Now  referring  to   Table 

@  8'  @>  g'  ©X^C®  9'  (fi'faeffisOh      A,  let  us  try  wheel  13  at  C. 

Then  the  loads  will  be  in  the 
position  shown  in  Fig.  195, 
and  for  the  average  unit  load 
on  the  left,  considering  half 


§J            <$l             $. 

\ 

/st-  o" 

/S-0"     F      /5-0" 

AS--<7" 

6O-O" 

Fig.   194 


1"    ,,      i 

15' 

ic   • 

IS' 

IS' 

r              rt 

I                                                      €0'-  0" 

Fig.   195 


of  wheel  13  as  being  on  the  left,  we  have 


73,000 
30 


=  2,430  Ibs.  (about), 


and  for  the  average  unit  load  on  the  whole  span,  we  have 

142,000 


60 


=  2,370  Ibs. 


Now  it  is  quickly  seen  that  this  position  will  give  the  maximum  moment 
at  C.  For  if  wheel  14  be  placed  at  C  there  would  be  80,000  Ibs.  to  the 
left  which  would  give  2,660  Ibs.  average  unit  load  to  the  left  and  the 
average  unit  load  on  the  span  would  be  the  same  as  given  above,  and  if 
wheel  12  be  placed  at  C  the  average  unit  load  on  the  left  will  be  53,000  -f 
30  =  1,760  Ibs.  and  the  average  unit  load  on  the  span  will  remain  the  same 
as  before.  So  the  wheels  when  in  the  position  shown  in  Fig.  195  will 
produce  the  maximum  bending  moment  on  the  girder. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  269 

Then,  the  next  thing  is  to  determine  the  bending  moment  about  wheel 
13  as  that  is  the  moment  desired.  To  do  this  we  must  first  determine  the 
reaction  R  at  A  which  we  do  by  taking  moments  about  B.  As  wheel  18 
just  happens  to  come  exactly  at  B  we  can  obtain  the  moments  of  all  the 
wheels  on  the  span  directly  from  Table  A.  Passing  down  the  line  through 
wheel  18  (in  the  table)  until  we  come  to  the  zigzag  line,  then  passing  to 
the  left  until  we  come  to  the  vertical  line  through  wheel  9,  we  find  the 
moment  4,224:  which  is  the  moment  in  thousands  of  foot  pounds  of  all  the 
wheels,  9  to  17  inclusive,  about  wheel  18,  which  happens  to  be  at  the  right 
support. 

Then  for  the  reaction  R  we  have 

70.4  (thousands). 


Then  taking  moments  about  wheel  13  we  have  (using  Table  A) 

M=  (70.4x30  -818)  1,000x12  =  15,528,000  inch  Ibs. 
Then  multiplying  this  by  50/40  to  reduce  it  to  Cooper's  £50,  we  have 

15,528,000x^-19,410,000  inch  Ibs. 

for  the  maximum  live-load  bending  moment  on  the  girder. 
Then  for  the  impact  we  have 


(  at?Q®~  } 

\  60  +  oOO  I 


19,410,000  x     at~      =  16,175,000  inch  Ibs. 


Now  adding  the  above  dead-  and  live-load  moments  and  impact  together 
we  have 

4,806,000  +  19,410,000  +  16,175,000  =  40,391,000  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  girder.     Now,  assuming 
the  web  will  be  f  "  thick,  we  have 


40,391,000       B-V. 
= 


for  the  economic  depth.     So  we.  will  use  a  web  84"  deep   (as  this  is  a 
common  plate)  and  assume  it  to  be  f  "  thick. 

The  total  depth  of  the  girder  back  to  back  of  angles  will  then  be 
84.25".  Then  subtracting  about  an  inch  we  have  83.2"  for  an  assumed 
effective  depth.  Dividing  this  into  the  maximum  moment  given  above  we 
have 

40,391.000  s   (about) 


for  the  flange  stress,  and  then  we  have 

486,000 
lg-656=  30.4  sq. 

for  the  required  net  area  of  the  cross-section  of  each  flange. 


486,000 

-=  30.4  sq.ms.  (about) 


270  STRUCTURAL  ENGINEERING 

The  area  of  the  cross-section  of  the  84"  x  $"  web  =  31. 5n",  and 
.J  =  3.9n".     Then  for  the  make-up  of  each  flange  we  have 

Gross      (J"  rivets)      Net 

2— Ls  6"  x  6"  x  ^"  =  15.56n"-2.75n"  =  12.80n" 

1— cov.  pi.  14"  x  f"  =    8.75n"-1.25n"  =    7.50n" 

1— cov.  pi.   14"  x  J"  =    7.00n"-1.00n"  =    6.00n" 

area  of  web  =    3.90n" 


31.31n"  30.20n" 

Now  taking  moments  about  the  center  of  the  top  cover  plate   (see 
Fig.  196)  we  have 

-     2.62  x  15.56  +  8.75  x  0.56 

x  =  —  ~~rnr~  -  =  1.46  ms. 

*  HR,  £>/  *<°  l>v  ^or  *^e  distance  from  the  center  of  the  top  cover 
^f1'"  ^  **  •-~t'cl  V  ^t  plate  to  the  center  of  gravity  of  the  whole  flange. 
5;L^t-2Lllp^5r.41l  Then  we  have 


1.46-0.87  =  0.59  ins. 

Fig.   196 

for   the    distance    from    the    back    of    the    flange 
angles  to  the  center  of  gravity  of  the  flange.     Then  we  have 

84.25 -(0.59x2)  =83.07  ins. 

for  the  actual  effective  depth  of  the  girder,  which  is  quite  close  to  that 
assumed  above,  so  no  recalculation  on  account  of  the  assumed  effective 
depth  is  necessary. 

For  the  length  of  the  y  cover  plates,  which  will  be  the  outside  cover 
plates  on  the  girder,  we  have 


=  60         =  28'-6"  +  1/-6//  =  30'- 


For  the  f  "  cover  plate  we  have 


y  '  =  60      rj     -  42'-6"  +  l'-6"  =  44'. 

The  maximum  end  shear  due  to  dead  load  is  equal  to  the  reaction 
shown  in  Fig.  194  plus  the  weight  of  the  girder  for  half  of  the  length  of 
the  end  panel.  The  four  flange  angles  of  the  girder,  as  given  above, 
weight  106*  per  ft.  of  girder,  the  web  weighs  108*  per  ft.,  and  the  weight 
of  stiffeners  and  details  at  this  point  on  the  girder  will  be  about  90  per 
cent  of  the  weight  of  the  web  per  ft.,  or  97*,  and  the  two  f  "  cover  plates 
weigh  about  60*,  making  in  all  about  370*,  say,  400*  per  ft.  of  girder. 
Then  we  have 

5  =  20,025  +  400x7.5  =  23,000  Ibs.  (about) 

for  the  maximum  end  shear  on  the  girder. 

The  maximum  end  shear  on  the  girder  due  to  live  load  is  equal  to 
the  maximum  shear  in  the  end  panel  due  to  that  load.  Now  according  to 
Art.  90  the  maximum  shear  in  the  end  panel  (the  same  as  in  a  truss 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  271 

bridge)  will  occur  when  the  load  in  the  panel  is  equal  to  the  total  load  on 
the  bridge  divided  by  the  number  of  panels.  So  the  first  thing  is  to 
satisfy  this  criterion  for  maxi- 
mum shear  in  the  end  panel.  . 
Placing  wheel  3  at  the  first  A@- 

panel  point  out  from  the  end  of        [5 — ^ — i ^ — i- — i 

the  girder  as  shown  in  Fig.  197        V G0  ~°" 

we  have  (using  Table  A)  Fig.  197 

+  20,000  +  10,000-40,000  Ibs. 


for  the  load  in  the  end  panel  AD,  considering  one-half  of  wheel  3  in  the 
panel.  With  the  wheels  in  this  position  the  total  load  on  the  span  is 
152,000*.  Then  we  have 

15*'000=  38,000  Ibs., 

4 

which  to  satisfy  the  criterion  should  be  40,000*.  But  this  is  as  close  as 
the  criterion  can  be  satisfied,  as  can  be  verified  by  trial. 

The  next  thing  is  to  determine  the  shear  in  the  end  panel  AD  with 
the  wheels  in  the  position  shown  in  Fig.  197.  First  of  all  the  shear  in  the 
panel  is  equal  to  the  reaction  R  at  A,  due  to  all  of  the  loads  on  the  span, 
minus  the  concentration  r  from  the  stringers,  due  to  the  loads  in  the  panel 
AD.  In  fewer  words,  the  shear  in  the  end  panel  AD  is 

S'  =  R-r. 

Taking  moments  about  the  right  support  B  (using  Table  A)  we  have 
fl=  4.638.000 +  158,000  x  2  =8g)3001hgj> 

and  taking  moments  about  the  panel  point  D  we  have 

230,000 

r  = — —  =  15,300  Ibs. 

15 

Then  for  the  maximum  shear  in  the  end  panel  AD  we  have 

S'  =  82,300  -  15,300  =  67,000  Ibs., 
and  multiplying  this  by  50/40  we  have 

67,000  x|g  =  84,000  Ibs.  (about) 

for  the  maximum  live-load  shear  on  the  girder. 
Then  for  the  impact  we  have 


Now  adding  together  the  dead-load  and  live-load  end  shears,  given  above, 
and  the  impact,  we  have 

23,000  +  84,000  +  73,000  =  180,000  Ibs. 


272  STRUCTURAL  ENGINEERING 

for  the  total  maximum  end  shear  on  the  girder,  which  is  practically  the 
shear  throughout  the  end  panel. 

Now  for  the  unit-shearing  stress  on  the  assumed  84"  x  £  "  web  we 


). 

Then  substituting  this  for  s  in  Formula  (1),  Art.  117,  we  have 
<J  =  A  (12,000  -5,710)  =59  ins. 

for  the  maximum  spacing  of  the  stiffeners  at  the  ends  of  the  girders. 
Now,,  as  this  is  not  less  than  the  half  depth  of  the  girder,  the  assumed 
web  is  satisfactory.  The  specification  will  not  permit  of  the  web  being 
made  thinner. 

The  outstanding  flanges  or  legs  of  the  stiffeners,  to  satisfy  the  speci- 
fications,, should  not  be  less  than  1/30x84  +  2  =  4.6".  This  requires  the 
use  of  5"  x  3J"  angles,  the  5"  leg  outstanding.  All  intermediate  stiffeners 
can  be  taken  as  5"  x  3J"  x  f"  angles  without  hesitating  —  this  being  the 
minimum  thickness  allowed  —  but  the  end  stiffeners  must  have  sufficient 
area  of  cross-section  to  take  the  maximum  reaction  on  the  girder  when 
considered  as  columns,  as  per  specifications.  The  cross-section  of  a  pair 
of  these  angles,  considered  as  a  column,  is  shown  in  Fig.  198. 


R 

15'         \ 

15' 

/5' 

\        ,s' 

60'- 

o" 

'9 

Fig.   198  Fig.   199 

Let  us  first  assume  them  to  be  5"  x  3J"  x  f"  angles.     Then  for  the 
radius  of  gyration  about  the  axis  y-y  we  have 


[2  x  7.78  +  2.48^  x  6.1  . 

r-  \| =2.96  ms. 

Then  taking  one-half  the  depth  of  the  girder  as  the  length,  as  per  speci- 
fication, we  have 

p  =  16,000  -  70  -^  =  15,000  Ibs.  (about) 

for  the  allowable  unit  stress  on  such  a  column. 

Now  from  Fig.  194  it  is  readily  seen  that  the  total  maximum  dead- 
load  reaction  is 

20,025  +  6,675  =  26,750  Ibs.,  say,  27,000  Ibs. 

The  maximum  live-load  reaction,  as  is  readily  seen,  will  occur  when 
the  wheels  are  in  the  position  shown  in  Fig.  199,  where  AB  represents  the 
span. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  273 

Taking  moments  about  B  (using  Table  A)  we  have 

f  162  x  4N 


_      /5,208  + 

R=(~  ~ 


1,000- 97,600  Ibs. 


for  the  maximum  live-load  reaction  at  A  due  to  Cooper's  £40,  and  multi- 
plying this  by  50/40,  to  reduce  it  to  Cooper's  £50  loading,  we  have 

97,600  x  jg  =  122,000  Ibs. 

for  the  maximum  live-load  reaction  desired. 
Then  for  the  impact  we  have 

=101,000  Ibs.  '.    '.    :'.-'      . 

Now  adding  together  the  above  maximum  dead-  and  live-load  reactions 
and  the  impact  we  have 

27,000  +  122,000  +  101,000  =  250,000  Ibs. 

for  the  total  maximum  reaction.     Now  dividing  this  by  the  allowable  unit 
stress  as  found  above  for  the  end  stiffeners  we  have 

250,000 

:  I6'6  sq<  ms' 


for  the  required  area  of  the  end  stiffeners. 

It  is  necessary  to  have  two  pairs  of  end  stiffeners  at  each  end  of  the 
girders  in  order  to  properly  distribute  the  pressure  over  the  pedestal. 
Two  pairs  of  the  stiffeners  assumed  above  have  12.  2n"  cross-section, 
which  is  about  4n/'  less  than  required.  So  thicker  angles  than  those 
assumed  will  have  to  be  used.  The  allowable  unit  stress  will  not  be 
materially  affected  by  using  thicker  angles  as  the  radius  of  gyration  varies 
but  little  with  the  thickness.  So  we  can  use  any  two  pairs  of  5"  x  3J" 
angles  that  have  the  proper  area.  It  is  seen  from  a  handbook,  or  from 
Table  4,  that  4—  Ls  5"  x  3J"  x  J"  have  16n"  cross-section,  and  that 
4—  Ls  5"  x  3£"  x  TV  have  17.88n".  As  is  seen,  the  J"  angles  have  an 
area  of  cross-section  that  is  0.6n"  smaller  than  required,  and  the  iV' 
angles  have  about  1.28n"  more  than  required.  So  let  us  use  4  —  Ls  5"  x 
3J"  x  \"  =  16.0n"  at  each  support  for  end  stiffeners. 

For  area  of  bearing  on  the  masonry  we  have  required 

250,000 


The  pedestal  should  be  made  to  suit. 

Preliminary  estimate  of  the  weight  of  one  main  girder. 

4—  Ls  6"  x  6"  x  ft"  x  69'-0"  x  26.5*  (combined 

with  end  Ls)  ..............................  7,314  Ibs. 

l_web  84"  x  f"  x  62'-0"  x  107.12*  .............  6,641  Ibs. 

1—  cov.  pi.  14"  x  J"  x  60'-0"  x  23.80*  (2  combined)  1,428  Ibs. 

1—  cov.  pi.  14"  x  f"  x  118'-0"  x  29.75*  ..........  3,510  Ibs.  18,893  Ibs. 


274 


STRUCTURAL  ENGINEERING 


8—  end  stiff.  Ls  5" 
26—  int.  stiff.  Ls  5" 


x  3J"  x  \"  x  13.6*  x  7'-0"  ......    762  Ibs. 

x  3J"  x  f"  x  10.4*  x  7'-0"  .....  1,893  Ibs. 


7—  fillers  8"  x  ii"  x  6'-0"  x  18.7*  (at  floor  beams)  .     785  Ibs. 


8—  splice  pis.  10"  x 
4—  splice  pis.  14 
3—  fills.  31"  x 


x  2'-3"  x  23.38*  ..........    421  Ibs. 

x          x  4'-4"  x  32.72*  .........    568  Ibs. 

x  6'-0"  x  8.18*  ...............    147  Ibs. 


4,576  Ibs. 


3%  for  rivet  heads 


23,469  Ibs. 
700  Ibs. 


Total  weight  of  one  main  girder 24,169  Ibs. 

Total  weight  of  two  main  girders 48,338  Ibs. 

Total  weight  of  main  girders  per  ft.  of  span  =  48,338/60  =  805  Ibs. 

To  make  this  estimate  the  student  would  have  to  sketch  the  details  of 
the  girders  to  some  extent. 

151.    Design  of  Lateral  System. — The    laterals    with    the    floor 
beams  and  main  girders  form  a  horizontal  double  truss  as  shown  in  Fig. 

200,  where  AB  represents  one  main  girder 
and  CD  the  other.  The  laterals,  according 
to  the  specifications,  must  be  designed  to 
resist  a  lateral  force  of  200  +  5,000x0.10 
o  -  700*  per  ft.  of  span  considered  as  a  live 
load.  One  system  can  be  considered  as 
taking  the  force  when  applied  from  one 
direction  and  the  other  system  when  it  is 
applied  from  the  other  direction.  So  the 


f 

F 

G 

^   / 

^     / 

\        / 

\       / 

X*-, 

X 

X 

X 

y   \ 

y  \ 

/       \ 

/      \ 

m    /s'      _[©   is' 

2 

(3)    fS' 

/               - 
/s 

t                4.  Panels®  IS-O"=  6O-  O  " 

Fig.   200 


laterals  may  be  considered  as  taking  tension  only. 
For  a  panel  load  we  have 

P  =  700x15  =  10,500  Ibs. 

Then  assuming  the  lateral  force  to  be  applied  from  the  direction  indicated 
by  the  arrows  and  suppose  it  moves  onto  the  structure  from  right  to  left, 
loading  panel  points  G  and  F,  we  have 

S  =  -x  10,500  =  7,800  Ibs. 
4 

for  the  maximum  shear  in  panel  FE  (see  Art.  90),  and  for  the  maximum 
shear  in  panel  EA,  we  have 

S'=-x  10,500  =  15,700  Ibs. 
4 

By  multiplying  each  of  these  shears  by  the  secant  of  the  angle  6  we  will 
obtain  the  stress  in  the  corresponding  laterals.  As  angle  0  is  about  45 
degrees  we  can  take  the  secant  as  1.4.  Then  for  the  stress  in  the  diagonal 
marked  (a)  we  have 

T  =  15,700x1.4  =  22,000  Ibs., 
and  for  the  stress  in  the  diagonal  marked  (6)  we  have 

r  =  7,800x1.4  =  10,900  Ibs. 

If  the  lateral  force  were  applied  in  the  opposite  direction  to  that 
indicated  by  the  arrows,  the  lateral  marked  (c)  would  be  subjected  to  a 
maximum  stress  of  22,000*  and  the  lateral  marked  (d)  would  be  subjected 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  275 

to  a  maximum  stress  of  10,900*  and  laterals  (a)  and  (6)  in  that  case 
would  have  no  stress.  The  same  maximum  stresses  would  be  found  in  the 
laterals  in  the  right  half  of  the  span  if  the  lateral  force  were  considered 
to  move  onto  the  span  from  left  to  right. 

The  stresses  in  the  floor  beams  due  to  the  lateral  force  (—  700*)  can 
be  neglected,  as  the  stress  produced  is  simple  compression  in  the  bottom 
flanges,  which  are  in  tension  from  dead  and  live  load.  It  is  readily  seen 
that  the  stresses  calculated  above  are  all  that  we  need  for  designing  the 
laterals. 

For  the  laterals  marked  (a)  and  (c)  we  have 

22,000 


for  the  required  area  of  cross-section. 

Use  1—  L  3£"  x  3"  x  f"  =  2.30  -  0.37  =  1.93°"  net.  This  is  the 
smallest  angle  allowed  by  the  specifications  and  therefore  the  same  size 
will  be  used  for  each  of  the  other  laterals. 

Preliminary  estimate  of  weight  of  lateral  system. 

8—  Ls  31"  x  3"  x  f"  x  7.9*  x  20'  .............  1,264  Ibs. 

6—  lat.  pis.  17"  x  f  "  x  21.6*  x  3'  .............  389  Ibs. 

4—  lat.  pis.  17"  x  |"  x  21.6*  x  2'  .............  173  Ibs. 

4—  sp.  pis.  9"  x  |"  x  11.5*  x  3'  ...............  138  Ibs. 

16—  Ls6"x4"x  i"  x  16.2*  x  1'  ...............  259  Ibs. 

rivets     ...................................  70  Ibs. 

2,293  Ibs. 

Total  weight  of  laterals  per  ft.  of  span  =  2,293/60  =  38*  per  ft. 
152.    Preliminary  Estimate  of  Weight  and  Cost.— 

Estimates  of  effective  dead  weight  per  ft.  of  span. 

Weight  of  stringers  per  ft.  of  span  (Art.  147)  .  .  289  Ibs. 
Weight   of   intermediate   floor   beam   per   ft.    of 

span  (3,621/15)    (Art.  148)  .  .  ...........  242  Ibs. 

Weight  of  main  girders  per  ft.  of  span  (Art.  150)  805  Ibs. 

Weight  of  laterals  per  ft.  of  span  (Art.  151)  ----  38  Ibs. 

Weight  of  deck  per  ft.  of  span  ................  400  Ibs. 

Total  weight  of  dead  load  per  ft.  of  span.  .  .    1,774  Ibs. 
As  1,780*  was  assumed  no  recalculations  are  necessary. 

Estimate  of  total  weight  of  metal  in  span. 

2  main  girders  at  24,169*  ...................  48,338  Ibs. 

3  intermediate  floor  beams  at  3,621*  .....  .  ----  10,863  Ibs. 

2  end  floor  beams  at  2,856*  .................    5,712  Ibs. 

4  panels  of  stringers  and  details  at  4,358*  .....  17,432  Ibs. 

8  laterals  and  details  ......  .................    2,293  Ibs. 

4  pedestals  and  sole  plates  ...................    2,200  Ibs. 

86,838  Ibs. 


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279 


280 


STRUCTURAL  ENGINEERING 


For  the  cost  at  3^  we  have  86,838  x  0.035  =  $3,039.33.  3J^  per  Ib. 
is  a  common  price  for  this  class  of  work  erected.  However,  the  pound 
price  will  vary  from  2-|^  to  4^.  It  depends  upon  the  market  price  of 
metal  and  freight. 

This  completes  the  necessary  preliminary  calculations  for  the  span. 
Next  a  stress  sheet  as  shown  in  Fig.  201  can  be  drawn.  Then  the  shop 
drawings  (Figs.  202  and  203)  for  the  span  can  be  made. 


Fig.   205 

The  work  up  to  the  completion  of  the  stress  sheet,  as  a  rule,  is  done 
in  the  designing  office  of  a  bridge  company,  or  in  the  office  of  a  railroad 
company,  or  in  the  office  of  a  consulting  engineer.  Consulting  engineers 
as  a  rule  (and  sometimes  railroad  companies)  make  general  drawings, 
similar  to  the  one  shown  in  Fig.  204,  instead  of  stress  sheets.  In  that 
case  the  general  drawing  for  the  span  is  submitted  to  the  different  bridge 
companies  for  use  in  preparing  their  bids  for  fabricating  and  (usually) 
erecting  the  structure.  After  the  contract  is  awarded,  the  general  draw- 
ing, instead  of  a  stress  sheet,  is  used  as  a  guide  in  making  the  shop  draw- 


DESIGN  OF  SIMPLE  EAILKOAD  BRIDGES  281 

ings  by  the  bridge  company  obtaining  the  contract.  As  a  rule  bridge 
companies  do  not  make  general  drawings  of  such  structures — only  the 
stress  sheets  and  shop  drawings. 

153.  Making  of  the  Detail  Drawings. — In  beginning  the  work, 
the  first  thing  to  do  is  to  draw  a  half  cross-section  of  the  span,,  as  shown 
in  Fig.  205,  to  a  large  scale,  say  1-|"  scale.  In  making  this  sketch  (Fig. 
205)  first  draw  the  cross-section  of  the  main  girder  and  the  stiffeners. 
Then  the  next  thing  to  do  is  to  draw  the  floor  beam.  Counting  from  the 
back  of  the  bottom  flange  angles  of  the  main  girder,  we  have  \^''  ^or  the 
thickness  of  the  flange  angle,  and  f "  for  the  thickness  of  the  lateral  plate, 
making  in  all  yi  +  t  =  Wif"  f°r  tne  distance  from  the  back  of  the  flange 
angles  of  the  main  girder  to  the  back  of  the  bottom  flange  angles  of  the 
floor  beam.  This  distance  being  determined,  the  bottom  flange  of  the  floor 
beam  is  located  and  can  be  drawn.  The  distance  from  the  back  of  the 
bottom  flange  angles  to  the  back  of  the  top  flange  angles  of  such  shallow 
floor  beams  is  usually  made  J"  more  than  the  depth  of  the  web.  So  in  this 
case,  as  the  web  is  32"  deep,  we  have  32 J"  for  the  distance  from  the  back 
of  the  bottom  flange  angles  to  the  back  of  the  top  flange  angles,  and  thus 
the  top  flange  is  located  and  can  be  drawn.  Next  the  cross-sections  of  the 
two  I-beams  composing  the  stringer  can  be  drawn  at  their  proper  location. 
Then  by  drawing  a  horizontal  line  9J"  above  the  top  of  these  beams 
(using  10"  ties  dapped  \"  on  the  stringer)  we  have  the  base  of  rail 
located,  and  having  the  base  of  rail  located  the  clearance  line  can  be  drawn 
as  shown.  This  is  about  as  far  as  we  can  proceed  with  the  drawing  (Fig. 
205)  until  the  required  spacing  of  the  rivets  in  the  floor  beam  is  deter- 
mined. 

For  the  shear  on  the  part  of  the  floor  beam  between  the  end  and 
where  the  stringer  connects,  we  have  137,200*  (see  stress  sheet).  The 
vertical  distance  from  the  center  of  rivets  in  the  top  flange  to  center  of 
rivets  in  the  bottom  flange  is  about  25.5",  and  the  allowable  bearing  of  a 
f"  shop  rivet  on  the  £"  web  is  10,500*.  So,  substituting  137,200  for  S, 
25.5  for  h,  and  10,500  for  r  in  Formula  (2)  (Art.  116),  we  have 

10,500  x  25.5 
P=        137,200         :1'95inS" 

say,  2",  for  the  theoretical  spacing  of  the  flange  rivets  in  the  part  of  the 
floor  beam  between  the  end  and  the  stringer. 

The  shear  on  the  part  of  the  floor  beam  between  the  stringers  is  very 
slight.  In  fact  there  is  no  shear  at  all  except  that  due  to  the  weight  of  the 
intervening  part  of  the  floor  beam  itself.  This  being  the  case,  the  flange 
rivets  in  that  part  of  the  floor  beam  can  be  spaced  6"  apart,  which  is  the 
maximum  spacing  allowed. 

Now  beginning  at  the  end  of  the  floor  beam  (Fig.  205)  we  space  the 
rivets  2"  apart  (as  required)  in  the  flange  angles  out  to  the  splice  (where 
the  gusset  and  web  meet)  and  the  only  break  in  the  2"  spacing  there  is 
that  necessary  to  fit  the  details  of  the  splice.  The  splice  is  located  so 
as  to  bring  the  bracket  just  inside  the  clearance  line  as  shown. 

The  rivets  in  the  splice  should  line  up  with  those  in  the  stringer 
connection  and  at  the  same  time  with  those  in  the  end  details  of  the  floor 
beam. 

There  must  be  enough  rivets  in  each  stringer  connection  to  transmit, 


282  STRUCTURAL  ENGINEERING 

in  bearing  on  the  web,  the  maximum  floor  beam  concentration.  For  the 
concentration  we  have  5,200*  from  the  dead  load,  68,200*  from  the  live 
load  as  given  in  Art.  148,  and  62,000*  [=300  -=-  (30  +  300)  x  68,200]  from 
the  impact,  making  a  total  of  135,400*.  The  allowable  bearing  of  a  I" 
field  rivet  on  the  \"  web  is  J  x  20,000  x  \  =>  8,750*.  Then  for  the  number 
of  rivets  required  in  each  stringer  connection  we  have 

135,400 

=  15.4,  say   16,  rivets. 


It  is  seen  from  this  that  a  single  line  of  rivets  in  each  of  the  angles 
connecting  the  stringers  to  the  floor  beam  is  sufficient.  So  we  have  this 
much  data  for  later  use. 

Now  the  next  thing  to  do  is  to  determine  the  vertical  spacing  of  the 
rivets  in  the  floor  beam.  It  appears  that  there  is  no  question  but  that  the 
rivets  should  be  spaced  as  closely  as  is  permissible  in  the  vertical  direction 
in  the  case  of  such  shallow  floor  beams.  So  we  will  simply  try  putting  in 
as  many  rivets  in  the  vertical  direction  as  will  fit  in.  Beginning  at  the 
bottom  of  the  floor  beam,  we  will  make  the  first  gauge  in  the  flange  angle 
2J",  as  the  angle  is  quite  thick  (yf"),  and  the  second  gauge  2|-",  which 
leaves  If  "  edge  distance  on  the  angle.  For  the  next  space  we  have  If" 
(edge  distance  on  the  angle)  plus  \"  (clearance)  plus  \\"  (edge  distance 
for  fillers  and  splice  plates),  making  in  all  3-J".  Now  these  same  spaces 
will  be  used  at  the  top  of  the  floor  beam  in  order  to  have  symmetrical 
spacing.  So  we  have 

2-f3      =17  ins. 


for  the  remaining  distance  in  which  rivets  are  to  be  spaced.  It  is  readily 
seen  that  3"  spacing  will  not  fit  in  this  distance  and  that  five  spaces  is  the 
maximum  number  possible.  So  we  will  space  the  rivets  as  shown. 

We  will  now  see  how  this  spacing  fits  the  detail  at  the  end  of  the  floor 
beam,  at  the  splice,  and  at  the  stringer  connection.  There  should  be 
enough  rivets  in  the  end  of  the  floor  beam  proper  to  transmit  the  greater 
part  of  the  maximum  end  shear  on  the  beam  to  the  main  girder.  In  the 
detail  shown  we  have  8  field  rivets  in  double  shear  or  bearing  on  lyV' 
metal.  The  double  shear  being  the  least  we  have  8  x  0.6  x  (10,000  x  2)  = 
96,000  Ibs.  for  the  amount  that  these  rivets  will  be  considered  to  transmit. 
This  leaves  41,200  Ibs.  (=137,200-96,000)  to  be  taken  by  the  rivets  in 
the  bracket  above  the  floor  beam  proper.  These  rivets  are  in  single  shear 
and  hence  the  allowable  stress  on  each  is  6,000  Ibs.  Then  we  have 

41,200 

•^55^=6.8  rivets,  say   7, 

required  in  the  bracket  above  the  beam  proper.  So  the  detail  as  shown  for 
the  end  of  the  floor  beam  is  quite  satisfactory,  as  the  extra  rivets  at  the  top 
of  the  bracket  are  really  needed  to  hold  the  bracket  which  stiffens  the  top 
flange  of  the  main  girder.  Now  by  prolonging  the  lines  giving  the  vertical 
spacing  on  to  the  stringer  connection,  it  is  seen  that  this  spacing  fits  nicely 
at  that  point,  as  we  obtain  the  detail  shown.  The  connecting  angles 
between  the  I-beams  are  shop  riveted  to  the  floor  beam  to  avoid  difficult 
field  riveting,  as  it  would  be  difficult  work  to  drive  the  rivets  connecting 
these  angles  to  the  floor  beams  in  the  field. 


DESIGN  OF  SIMPLE  EAILROAD  BEIDGES  283 

Let  us  now  turn  our  attention  to  the  splice.  The  most  satisfactory 
way  of  designing  such  splices  is  to  draw  in  what  looks  to  be  about  correct, 
and  then  determine  the  stress  on  the  rivets  assumed,  and  make  whatever 
modifications  are  necessary.  So  let  us  assume  we  need  three  rows  of  rivets 
on  each  side  of  the  splice  as  shown.  First  suppose  that  the  f"  plates 
extending  over  the  flange  angles  are  omitted  and  that  there  is  just  a  single 
f "  splice  plate  on  each  side  of  the  web. 

Now,  for  the  vertical  force  on  each  rivet  due  to  the  vertical  shear  we 
have 

137,200 
v  = — —  =7,620  Ibs., 

and  let  *  be  the  maximum  horizontal  force  on  each  of  the  rivets  farthest 
out  from  the  center  of  the  web  due  to  cross  bending.  Then,  according  to 
Art.  117,  we  have  the  equation 


2x3(1.75    +5.25  +8.5  )  xs-f  8.5  =  1/8  area  of  web 

multiplied  by  16,000  x  h  (=2  x  16,000  x  29)  =928,000  in.  Ibs.,  from  which 
we  obtain 

,=  928.000  =  lbs 

72.6 

This  horizontal  force  should  not  exceed  ($•  x  24,000  x  £)  17/25.5  =  7,000 
Ibs.,  so  the  splice  as  just  considered  would  not  be  sufficient. 

As  one  more  line  of  rivets  on  each  side  of  the  splice  will  not  be 
sufficient,  it  is  obvious  that  it  will  be  best  to  use  the  f"  plates  extending 
over  the  flange  angles  as  shown.  So,  considering  the  splice  as  it  is  shown 
in  Fig.  205,  we  have 


=  928,000  in.  Ibs., 

from  which  we  obtain 

928,000  _  „      ,  ,      .  v 

1390    =6'y°°  lbs<  (about) 

for  the  horizontal  force  on  the  outer  rivets  (which  are  those  in  the  outer 
gauge  lines  in  the  flange  angles).  These  rivets  are  so  near  the  top  and 
bottom  edge  of  the  beam  that  the  vertical  shear  on  them  can  be  neglected 
(see  Art.  61). 

The  above  6,700*  force  can  be  considered  as  applied  to  the  rivets  by 
the  web  and  transmitted  from  there  to  the  angles  and  on  to  the  f"  plates 
producing  a  shear  of  3,330*  on  each  rivet  at  each  f"  plate.  The  flange 
increment  can  be  considered  as  being  transmitted  by  the  f"  plates  and 
web  combined.  Taking  the  flange  increment  as  10,500*  (allowable  bear- 
ing of  a  J"  rivet  on  the  J"  web)  and  assuming  that  one-third  is  trans- 
mitted to  the  flange  angles  by  the  web  and  one-third  by  each  of  the  f" 
plates,  we  have 

+  3,330  =  6,830  Ibs. 


284  STRUCTURAL  ENGINEERING 

for  the  shear  on  each  rivet  at  each  f  "  plate  and 
10,500 


for  the  bearing  stress  on  the  web.  Now,  as  7,200*  is  allowed  in  the 
former  case  and  10,500*  in  the  latter,  the  top  and  bottom  part  of  the 
splice  is  about  correct. 

The  rivets  connecting  the  splice  plates  to  the  web  directly,  not  passing 
through  the  flange  angles,  are  in  double  shear  and  bearing  on  the  web. 
The  allowable  bearing  on  the  web,  which  is  10,500  Ibs.,  is  less  than  double 
shear,  and  hence  the  maximum  stress  on  these  rivets  results  from  the 
bearing  on  the  web.  Now,  as  a  test  for  these  rivets,  let  us  consider  the 
ones  farthest  out  from  the  center  of  the  web,  which  are  8^"  out.  Let 
s"  be  the  horizontal  force  on  each.  Then  we  have 


2  I  3  (L752  +  57252+  8J5*)  +2  (TTTif  +  13.752)J^  =928,000  in.  Ibs., 


from  which  we  obtain 

„     928,000 
s"=  —  —  —  —  =  4,140  Ibs.  (about) 


(this  could  be  as  great  as  7,000  Ibs.). 

Then  for  the  resultant  force  on  each  of  these  rivets  we  have 


=  8,670  Ibs., 

which  is  less  than  the  allowable  bearing  of  each  on  the  \"  web,  and  hence 
the  splice  as  shown  is  amply  strong. 

The  splice,  as  shown  in  Fig.  205,  is  designed  to  develop  the  strength 
of  the  beam  (as  should  be  done),  while  as  a  matter  of  fact  the  flange  at 
the  point  of  splice  is  practically  twice  as  strong  as  need  be,  as  the  bending 
moment  at  that  point  is  only  about  one-half  the  maximum  on  the  beam 
and  if  the  splice  were  designed  to  carry  only  the  shear  no  actual  weakness 
would  be  detected  in  it. 

This  completes  the  necessary  calculation  in  connection  with  the 
drawing  of  the  general  cross-section  shown  in  Fig.  205.  Next  a  sketch 
to  1J"  scale  for  each  of  the  lateral  connections,  and  also  a  sketch  (to  a 
large  scale)  of  the  curved  end  of  the  main  girder,  should  be  made.  After 
these  preliminary  sketches  are  made,  we  can  proceed  with  the  making  of 
either  the  general  drawing  (Fig.  204)  or  the  shop  drawings  (Figs.  202 
and  203).  We  simply  transfer  the  dimensions  on  these  sketches  to  the 
final  drawings. 

Most  of  the  calculations  for  the  details  shown  on  these  drawings  are 
quite  similar  to  those  given  in  Art.  135  for  the  50-ft.  deck  plate  girder. 
The  main  difference  is  in  the  determination  of  the  flange  rivets  in  the 
main  girders.  In  the  case  of  the  through  girder,  here  considered,  the 
loads  are  applied  directly  to  the  web  of  the  main  girders  (by  the  floor 
beams)  instead  of  to  the  flanges  as  in  the  case  of  the  deck  girder  treated 
in  Art.  135. 

So  Formula  (2)  of  Art.  116  is  used  to  determine  the  pitch  of  the 
flange  rivets  instead  of  Formula  (4).  The  shear  in  each  panel  is  prac- 
tically constant  throughout  the  panel,  and  consequently  the  rivet  spacing 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  285 

should  be  constant  throughout  each  respective  panel.  Taking  the  case  of 
the  end  panel,  we  have 

S=  180,000*   (maximum  end  shear); 

r  =  7,880*  (=  allowable  bearing  of  a  £"  rivet  on  the  f"  web); 
ft  =  77". 

Now  substituting  these  values  in  Formula  (2)    (Art.  116),  we  have 

7,880  x  77 
P=    180,000    =3.36  ins.,  say,  3i  ins., 

for  the  theoretical  pitch  of  the  flange  rivets  in  the  end  panels.  The  shear 
on  the  girder  is  reduced  some  by  the  weight  of  the  girder  itself,  as  we 
pass  from  the  end  toward  the  center,  and,  consequently,  the  pitch  can  be 
increased  slightly  as  the  first  intermediate  floor  beam  is  approached,  as  is 
shown  in  Fig.  202. 

The  dead-load  shear  in  the  second  panel  from  the  end,  as  seen  from 
Fig.  194,  Art.  150,  is 

20,025-13,350  =  6,670  Ibs. 

The  maximum  live-load  shear  occurs  in  the  second  panel  when  the 
wheel  loads  are  in  the  position  shown  in  Fig.  206,  as  the  criterion  of  Art. 
90  is  satisfied.  / 

Taking  moments   about          |r       Os'  ©s®s®s(s)  9'  ©5'®  » 


the  end  B  of  the  span  (Fig.         TR       ,         \°  \c 

OA/»\  1.  /       •  m  IS  A          IS  -I-  is'         J.  IS' 

206)    we    have    (using    Ta-  6g;_o//         •*• 


ble  A)  Fig.  206 

2,155,000 


for  the  reaction  at  A.     Then  taking  moments  about  the  floor  beam  at  C 
we  have 

=5,330  Ibs.     ..  .  : 


(_    . 

for  the  stringer  concentration  on  the  floor  beam  at  D.  Then  we  have 
R  -  r  =  37,900  -  5,330  =  32,570  Ibs.  for  the  live-load  shear  in  the  second 
panel  from  the  end  of  the  span  due  to  Cooper's  £40;  and  for  Cooper's 
£50  we  have 


32,570x|5  =  40,700  Ibs., 


and  for  impact  we  have 


Now  adding  the  above  dead-  and  live-load  shears  and  the  impact 
together,  we  have 

6,670  +  40,700  +  37,000  =  84,370  Ibs. 

for  the  total  maximum  shear  in  the  second  panel  from  the  end  of  the  span, 
Then  using  Formula  (2)  (Art.  116)  we  have 

7,880  x  74    R  Q  . 

^.9  ms.  (about) 


286 


STRUCTURAL  ENGINEERING 


for  the  theoretical  spacing  of  the  flange  rivets  in  the  main  girders  through- 
out the  second  panel  from  the  end  of  the  span.  So  6",  the  maximum 
spacing  allowed,  will  be  used. 

Further  analyses  of  the  details  of  the  above  span  are  considered 
unnecessary,  for  if  the  details  of  the  deck  plate  girder  spans  previously 
treated  are  thoroughly  understood,  the  student  should  have  little  trouble  in 
making  the  detail  drawings  for  through  plate  girder  spans,  and  especially 
if  the  drawings  shown  in  Figs.  202,  203,  and  204  be  observed  closely  as 
the  work  progresses.  The  shop  bills  for  this  work  are  practically  the 
same  as  for  deck  spans. 

154.  General  Remarks. — The  general  discussion  given  above, 
regarding  the  design  of  flanges  and  end  bearings  for  deck  plate  girder 
bridges,  holds  in  the  case  of  through  plate  girder  bridges. 

The  floor  beams  for  through  plate  girder  bridges  are  sometimes  made 
as  shown  in  Fig.  207.  This  type  of  detail  eliminates  the  web  splice  in  the 


Fig.   207 

floor  beam.  The  brackets  are  usually  shipped  loose  and  riveted  to  the 
floor  beams  and  main  girders  in  the  field,  especially  when  the  main 
girders  are  quite  deep.  While  this  type  simplifies  the  details  of  the  floor 
beam,  it  complicates  the  details  of  the  main  girders  and  there  are  con- 
siderably more  field  rivets  than  there  are  in  the  type  used  above  for  the 
60-ft.  span.  The  details  of  the  two  types  vary  considerably  in  practice, 
and  the  details  given  here  are  intended  only  as  a  fair  example. 

The  stringers  for  through  plate  girder  bridges  are  sometimes  plate 
girders  (where  the  under  clearance  will  permit),  in  which  case  two 
stringers  per  track  are  used. 

155.  Graphical  Determination  of  Live-Load  Shears  and  Bend- 
ing Moments  on  Main  Girders  of  Through  Plate  Girder  Bridges.— 
The  "influence  line"  method,  outlined  in  Arts.  100  and  101,  is  the  most 
convenient  graphical  method  in  the  case  of  through  plate  girder  bridges. 
However,  the  equilibrium  polygon  can  be  used  if  one  so  desires. 

As  an  illustration,  let  it  be  required  to  determine  the  maximum 
reaction,  end  shear,  and  bending  moment  on  a  75-ft.  single-track  through 
plate  girder  span  of  five  15-ft.  panels,  due  to  Cooper's  £50  loading.  (See 
diagram,  Fig.  151.) 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES 


287 


To  determine  the  maximum  reaction,  lay  off  the  span  AB  (Fig.  208") 
to,  say,  i"  scale,  and  place  the  loads  as  shown. 

Then  draw  the  base  line  ab  arid  lay  off  ac  =  V  —  1*  and  draw  the 
influence  line  cb,  and  draw  the  ordinates  1,  2,  ...  13  under  the  loads,  as 
shown,  and  all  is  ready  for  determining  the  reaction.  Take  a  pair  of 
dividers  and  step  off  the  ordinates  1  to  8  as  explained  in  Example  1,  Art. 
100,  and  multiply  their  sum,  in  inches,  by  25,000*.  Next,  in  the  same 
way,  step  off  the  ordinates  9  to  12  and  multiply  their  sum  by  16,250*. 


Fig.   208 


Then  multiply  the  length  of  ordinate  13  by  12,500  and  add  all  three  of 
the  results  together,  and  the  desired  reaction  is  thus  obtained. 

To  determine  the  maximum  shear  in  the  end  panel  AD  (Fig.  209), 
place  the  wheels  as  shown,  thus  satisfying  the  criterion  for  maximum 
shear  (Art.  90).  Then  draw  the  reaction  influence  line  cb  for  the  span 
and  cd  for  the  panel  AD.  Then  by  multiplying  the  sum  of  the  ordinates 
(obtained  as  explained  above)  ef,  1,  2,  3,  4,  and  5  by  25,000*,  6  to  9  by 
16,250*,  and  h  and  10  by  12,500*,  and  adding  these  products  together, 
the  maximum  shear  in  the  end  panel  AD  is  obtained. 

To  determine  the  maximum  shear  in  any  of  the  intermediate  panels 


(/)     8    (^£(3)5^(5}     9'  (6)5  (7)6  (8)£(9)   S'  (fo  8    (//)*(/ 


Fig.   209 


as  DE  (Fig.  210),  first  draw  the  reaction  influence  lines  eb  and  af  and 
then  the  influence  line  adcb  for  shear  in  the  panel,  as  explained  for  trusses 
in  Art.  102,  and  place  the  wheels  as  shown  for  maximum  shear  in  the 
panel,  as  per  Art.  90.  Then  by  multiplying  the  sum  of  the  ordinates 
2  to  5  by  25,000*,  6  to  9  by  16,250*,  and  ordinate  1  by  12,500*,  and 
adding  the  products  together,  the  maximum  shear  in  the  panel  DE  is 
obtained. 

It  should  be  noted  in  the  last  case  that  the  position  of  the  wheels  for 
maximum  shear  in  the  panel  can  be  determined  directly  from  the  influence 
line.  For,  according  to  Art.  102,  no  load  should  pass  the  point  0  (Fig. 
210),  and  there  must  be  a  load  at  E,  so  by  placing  wheel  1  as  near  0  as 
possible,  with  a  wheel  at  E,  we  have  the  position  of  the  wheels  for 


268 


STRUCTURAL  ENGINEERING 


maximum  shear  in  the  panel.  The  position  of  the  wheels  for  maximum 
shear  in  any  of  the  intermediate  panels  can  be  determined  in  this  manner. 
The  maximum  bending  moment  will  undoubtedly  occur  at  the  floor 
beam  nearest  the  center  of  the  span,  and,  as  there  are  two  floor  beams 
live  load  (in  reference  to  C)  as  shown,  thus  satisfying  the  criterion  for 


Fig.   210 

equally  near  the  center,  the  location  of  either  beam  can  be  taken  as  the 
point  of  maximum  moment.  So,  if  AB  (Fig.  211)  represents  the  span, 
either  C  or  D  can  be  taken  as  the  point  of  maximum  bending  moment. 
Let  us  take  point  C.  Then  to  determine  the  maximum  moment,  place  the 
maximum  moment  given  for  trusses  in  Art.  91.  Next  construct  the 
influence  line  aOb  for  the  bending  moment  at  C  as  explained  in  Art.  101, 
and  also  in  Art.  102.  Then  by  multiplying  the  sum  of  the  ordinates  1  to  4 
by  25,000#,  5  to  9  by  16,250#,  10  by  12,500*,  and  one-half  of  11  by 


Fig.  211 

(2,500*. x  10')  and  adding  these  products  together,  we  obtain  the  maxi- 
mum bending  moment  on  the  main  girders. 

To  determine  the  maximum  shear  in  any  panel,  as  CD  (Fig.  212), 
by  means  of  an  equilibrium  polygon,  first  place  the  load  for  maximum 
shear  in  the  panel,  as  shown,  and  draw  the  ray  diagram  MNO,  and  the 
corresponding  equilibrium  polygon  a-c-b  .  .  .  a.  Then  by  drawing  OE, 
in  the  ray  diagram,  parallel  to  the  closing  line  ac  we  have  the  reaction 
(R)  at  A  due  to  the  loads  given  by  the  line  EM.  The  shear  in  panel  CD 
is  equal  to  R  minus  the  stringer  reaction  (r)  at  C  due  to  the  load  in  panel 
CD.  By  drawing  the  closing  line  ek  we  have  the  equilibrium  polygon 
eJche  for  the  panel  CD.  Then  by  drawing  E'O  (in  the  ray  diagram) 
parallel  to  the  closing  line  eJc  we  have  the  stringer  reaction  r,  at  C,  due  to 
wheel  1  (as  that  is  the  only  load  in  the  panel),  given  by  the  line  E'M. 
Then,  evidently,  the  maximum  shear  in  panel  CD  is  given  by  the  line  EE'. 


DESIGN  OF  SIMPLE  KAILEOAD  BEIDGES 


289 


The  maximum  shear  in  the  other  panels  can  be  determined  in  the  same 
manner. 

To  determine  the  maximum  bending  moment  on  the  main  girders  by 
means  of  an  equilibrium  polygon,  first  place  the  loads  on  the  span  for 


Fig.   212 


maximum  moment  at  any  panel  point  as  C  (Fig.  213),  as  explained  above, 
and  draw  the  ray  diagram  STP  and  the  corresponding  equilibrium  polygon 
acda.  Then  the  maximum  moment  at  C  is  equal  to  the  ordinate  y  multi- 
plied by  the  pole  distance  H.  H  is  measured  in  pounds  to  the  same  scale 
as  used  in  drawing  the  ray  diagram  and  y  is  measured  in  feet  or  inches  to 


Fig.   213 


the  same  scale  as  used  in  drawing  the  span  AB  and  in  spacing  the  loads. 
If  y  is  taken  in  feet  the  moment  will  be  in  foot  pounds,  and  if  taken  in 
inches  the  moment  will  be  in  inch  pounds. 

In  case  an  equilibrium  polygon  similar  to  the  one  shown  in  Fig.  186 
(Art.  144)  be  drawn  the  shears  and  bending  moments  for  any  number  of 
through  plate  girder  spans  can  be  determined  from  it. 


290 


STRUCTURAL  ENGINEERING 


The  shear  would  be  determined  as  shown  in  Fig.  212,  the  wheels 
being  placed  for  maximum  shear  as  per  criterion,  Art.  90.  The  maximum 
moment  at  any  panel  point  C  can,  however,  be  determined  without  refer- 
ence to  the  criterion  for  maximum  bending  moment,  very  much  the  same 
as  the  case  of  deck  spans  treated  in  Art.  144,  the  main  difference  being 
that  the  moment  here  is  for  a  certain  point  on  the  girder. 

As  an  illustration  let  UVW  (Fig.  214)  represent  an  equilibrium 
polygon,  similar  to  the  one  shown  in  Fig.  186.  We  start,  say,  by  first 
placing  the  point  C,  the  panel  point  where  the  moment  is  desired,  under 


wheel  11  and  drawing  the  closing  line  1-1,  we  obtain  the  ordinate  y.  Next 
placing  the  point  C  under  wheel  12  and  drawing  the  closing  line  2-2  we 
obtain  the  ordinate  y' ,  and  placing  C  under  wheel  13  and  drawing  the 
closing  line  3-3  we  obtain  the  ordinate  y" ,  and  so  on.  In  this  manner  the 
maximum  ordinate  for  point  C  is  obtained  by  trial,  and  by  multiplying 
this  maximum  ordinate  by  the  pole  distance  the  maximum  bending  moment 
is  obtained. 

The  maximum  bending  moment  at  any  panel  point  can  be  obtained  in 
this  manner. 

156.    Plate  Girder  Bridges  with  Solid  Floors.  —  Plate  girder 

bridges,  especially  through  spans,  sometimes  have  solid  floors,  particularly 
those  over  streets  in  cities  where  an  ordinary  open  floor  is  undesirable 
owing  to  the  dripping  of  water  from  the  floor  upon  the  sidewalk  and 
street  below  due  to  rain  and  melting  snow. 

These  structures,  in  reference  to  floor,  taking  the  most  common  con- 
struction, can  be  divided  into  two  general  types :  those  having  solid  metal 
floors  covered  with  concrete  and  ballast,  and  those  having  reinforced  con- 
crete floors  covered  with  ballast.  The  most  common  of  the  solid  metal 
floors  is  that  known  as  the  trough  floor  shown  in  Fig.  215  where  the  floor 
proper  is  made  up  of  vertical  and  horizontal  plates  connected  by  angles 
so  as  to  form  successive  troughs.  In  designing  such  bridges  the  first  thing 
to  do  is  to  draw  a  longitudinal  section  of  the  floor,  enough  to  include  three 
ties  18"  on  centers,  as  shown  to  a  large  scale  in  Fig.  215.  At  the  start  the 
whole  thing  is  an  assumption,  that  is,  we  draw  in  what  looks  to  be  correct. 


291 


292 


STRUCTURAL  ENGINEERING 


From  this  sketch  we  determine  the  dead  weight  of  the  floor,  including 
metal,  concrete,  ballast  and  track.  Then  we  compute  the  dead-  and  live- 
load  stresses  on  this  assumed  floor  and  if  necessary  modify  the  design  until 
it  agrees  with  the  final  computation. 

As  an  example  let  it  be  required  to  design  the  through  plate  girder 
span  indicated  in  Fig.  215,  using  Cooper's  E50  loading.  Estimating  from 
the  longitudinal  section  of  the  floor  (Fig.  215),  taking  concrete  to  weigh 
150*  and  ballast  125*  per  cu.  ft.,  we  have  for  a  strip  through  the  floor  one 
foot  wide  and  two  feet  long,  along  the  track,  the  following  weights : 

H  cu.  ft.  of  concrete  at  150* 225  Ibs. 

TCU.  ft.  of  ballast  at  125* 125  Ibs. 

Tie  (6  ft.  B.  M.)  at  6x4^* 27  Ibs. 

31  ft.  of  12"  x  f"  plates  at  15.3*.  .  53  Ibs. 

4  "ft.  3Jx3ixf  Ls  at  8.5* 34  Ibs. 

Rails  (90*)  and  spikes 12  Ibs. 

Tar,  stone  dust,  and  burlap 20  Ibs. 

Total 496  Ibs. 

for  two  square  feet  of  floor,  or  248*  for  each  square 
foot  (horizontal  projection).  The  floor  can  be  con- 
sidered as  being  made  up  of  independent  Z-shape  sec- 
tions, each  composed  of  one  vertical  plate,  two  angles 
and  two  halves  of  horizontal  plates  as  indicated  in 
Fig.  216  Fig.  216.  This  section  can  be  treated  as  an  inde- 

pendent transverse  beam.      For  the  moment  of  inertia 
of  this  beam  with  reference  to  axis  0-0,  we  have 


(43 


6"  x  f"  .............     =186   (hor.pls.) 

.....................   =       54   (vert.pl.) 


2  (5T25"  x  5  +  2.87)  ..............   =  281    (angles) 

521 
~(1.5n"  x  6J25*)  (rivets  staggered)  ..        -60 


Total 


461 

'   I2'@248*perlin,ft. 


111 


IS-  6' 


Fig.   217 


If  the  main  girders  are  15'-6"  apart 
the  dead  load  on  our  Z-beam  will  really  be 
as  indicated  in  Fig.  217,  as  the  248-lb.  load 
computed  above  holds  only  for  the  part  of 
the  floor  directly  under  the  track.  But  as 
the  final  result  will  not  be  affected  mate- 
rially if  we  assume  the  248  Ibs.  to  extend 
the  full  length  of  the  beam,  we  shall  so  consider  it. 

Then  for  the  maximum  bending  moment  due  to  dead  load  we  have 

M  =  J  x  248  x!5^52x  12  =  89,370  incVi  Ibs. 

The  next  thing  is  to  determine  the  maximum  bending  moment  due  to 
live  load.      In  accordance   with  the   usual  practice  we  will  assume  the 


DESIGN  OF  SIMPLE  EAILKOAD  BEIDGES  293 

maximum  live  load  in  this  case  to  be  the  heaviest  axle,  distributed  over 
three  ties.     So  we  have  (special  load,  see  diagram,  Fig.  151) 

'  =  20,833  Ibs. 


for  the  total  load  on  each  tie.  Now  from  the  longitudinal  section  of  the 
floor  (Fig.  215)  we  can  see  that  our  Z-beams  carry  from  half  of  this  load 
to  practically  all  of  it.  So  to  be  sure  we  will  assume  that  each  carries  the 
20,833  Ibs.  uniformly  distributed  along  the  beam  by  the  tie. 

Then  for  the  maximum  bending  moment  due  to  live  load  we  have 

M'  =  4  x  20,833  x  15.5  x  12  =  484,300  inch  Ibs. 

Now  adding  together  the  above  dead-  and  live-load  moments  and  allowing 
100  per  cent  for  impact  we  have 

M"  =  89,370  +  2  (484,300)=  1,057,970  inch  Ibs. 

for  the  total  maximum  bending  moment  on  our  Z-beam. 

Then  substituting  the  above  values  in  Formula  (D),  (Art.  53),  we 
have 

/=  1,057,970  x  6.6  =  1 

for  the  maximum  bending  stress  on  our  Z-beam  which  is  the  maximum 
bending  stress  on  the  floor.  This  shows  that  the  floor  is  but  little  heavier 
than  necessary  for  cross  bending,  16,000  Ibs.  being  the  allowable  bending 
stress. 

For  the  end  shear  on  our  Z-beam  we  have 

248x^=1,920  Ibs. 


from  dead  load  and 

20,833 


=  10,416  Ibs. 


from  live  load  and  the  same  from  impact,  making  22,700  Ibs.  in  all. 
Then  .to  resist  this  shear  we  should  have 

22,700 

16^00  =*•**  •*«"•. 

of  cross-section  in  the  vertical  plate  of  our  Z-beam.  This  shows  that  the 
shear  on  the  floor  is  amply  provided  for,  as  each  vertical  plate  contains 
4.5  sq.  ins. 

The  above  calculations  show  that  the  floor  shown  in  Fig.  215  is  about 
as  correct  as  we  can  design  it,  as  f  "  metal  is  the  minimum  thickness 
allowed. 

Having  the  floor  designed,  the  dead  weight  coming  onto  the  main 
girders  from  the  floor  can  be  computed  and  by  adding  this  per  foot  of 
girder  to  the  assumed  weight  of  girder  itself  (per  ft.)  we  obtain  the 
dead  load  for  determining  the  dead-load  stress  in  the  main  girders.  Then, 
using  the  formula  pL2/8,  the  maximum  dead-load  bending  moment  on  the 
main  girders  can  be  determined. 


294 


STRUCTURAL  ENGINEERING 


To  obtain  the  maximum  live-load  shear  and  bending  moment  on  the 
main  girders  the  live  load  is  placed  just  the  same  as  in  the  case  of  deck 
spans. 

The  trough  floor  shown  in  Fig.  215,  while  it  is  the  most  common  type 
and  a  good  design,,  is  only  one  of  several  types  of  solid  floors  in  use.  For 
example,,  there  is  the  buckle  plate  floor  composed  of  buckle  plates  (see 
manufacturers'  handbooks)  laid  upon  I-beam  or  plate  girder  stringers, 
flat  plates  laid  upon  I-beams  used  either  as  stringers  or  transverse  beams, 
the  rolled  trough  section,  etc.  In  addition  to  these  types  there  are  the 
I-beam  and  concrete  floors  shown  in  Fig.  218,  in  which  case  the  I-beams 
are  designed  to  carry  all  of  the  load.  All  of  these  types  are  used  mostly 

rBollasl 


•Ballast 

Concrete, 


Fig.  218 


Fig.  219 


for  through  plate  girder  bridges.-  The  trough  floor  is  sometimes  used  on 
deck  spans,  in  which  case  the  troughs  are  placed  transversely  resting 
directly  upon  the  main  girders. 

The  out  and  out  reinforced  concrete  floors  are  used  on  deck  plate 
girders.  These  floors  are  simple  concrete  slabs  resting  directly  upon  the 
top  of  the  main  girders  and  turned  up  at  each  end  so  as  to  hold  the 
ballast,  as  shown  in  Fig.  219. 

In  designing  all  plate  girder  bridges  having  solid  floors,  the  first 
thing  to  do  is  to  design  the  floor  and  determine  its  weight.  Then  the 
weight  of  the  main  girders  can  be  assumed  and  added  to  the  weight  of  the 
floor  and  we  have  the  dead  weight  for  determining  the  dead-load  stresses 
in  the  main  girders.  The  other  work  involved  is  practically  the  same  as 
given  above  for  ordinary  bridges. 

157.  Double-Track  Spans. — Double-track  deck  plate  girder  bridges 
are  practically  always  composed  of  two  simple  single-track  spans  placed 
side  by  side,  as  previously  stated.  Double-track  through  spans  are  usually 
composed  of  two  main  girders  placed  far  enough  apart  to  admit  the  two 
parallel  tracks  13-ft.  centers.  In  the  case  of  ordinary  open  floors  there 
are  usually  four  lines  of  stringers.  The  load  in  that  case  on  each  stringer 
is  just  the  same  as  in  the  case  of  single-track  spans,  while  the  floor  beams 
have  four  equal  concentrations,  each  concentration  being  the  same  as  for  a 
single-track  span.  The  dead  load  on  the  main  girders,  as  stated  in  Art. 
124,  is  about  70  per  cent  more  than  for  single-track  spans,  while  the  live 
load  is  just  twice  as  much,  two  trains  being  considered  as  moving  abreast 
over  the  structure.  In  the  case  of  double-track  through  plate  girder 
bridges  composed  of  three  main  girders,  there  are  two  independent  single- 
track  floor  systems.  The  loads  in  that  case  carried  by  each  floor  and  by 
each  outside  main  girder  are  the  same  as  for  a  single-track  span,  while  the 
central  girder  carries  practically  twice  as  much  as  each  of  the  outside 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


295 


girders.  The  objection  to  the  three-girder  type  is  that  it  is  usually 
necessary  to  "spread"  the  tracks  at  the  location  of  these  bridges  in  order 
to  obtain  the  necessary  side  clearance. 

DRAWING  ROOM  EXERCISE  NO.  5 

Design  a  64-ft.  single-track  through  plate  girder  bridge  and  make  a 
stress  sheet  for  same  upon  a  24"  x  IS"  sheet  and  tracing  of  same. 

Length  of  span  =  4  panels  at  16'-0"  =  64'-0"  c.c.  end  bearings. 

Width  =  15'-S"  c.c.  girders. 

Live  load,  Cooper's  E50  loading. 

Dead  load,  to  be  assumed  by  student. 

Specifications,  A.  R.  E.  Ass'n. 

The  required  work  consists  of  making  the  calculations  and  the 
drawing  of  a  stress  sheet  for  the  span,  similar  to  the  one  shown  in  Fig. 
201  for  the  60-ft.  span. 

VIADUCTS 

158.  Preliminary. — The  ordinary  railroad  viaduct  is  a  bridge  com- 
posed of  a  series  of  deck  plate  girder  spans  supported  upon  towers.  How- 
ever, as  a  rule  any  bridge  supported  upon  towers  is  classed  as  a  viaduct. 
Viaducts  are  used  where  the  crossings  are  so  deep  that  ordinary  concrete 
or  masonry  piers  are  impracticable  as  to  cost.  The  most  common  type  of 
viaduct  is  shown  in  Fig.  220 (a).  The  type  shown  in  Fig.  220(6)  is 
built  to  some  extent.  The  principal  difference  in  the  two  types  is  in  the 


Cross  Section 


(a} 


Cross  Section 


Fig.  220 


tower  bracing.  Two  columns  connected  together  transversely  by  bracing 
constitutes  what  is  known  as  a  bent  (see  Fig.  221).  Two  bents  connected 
together  by  longitudinal  bracing  constitutes  what  is  known  as  a  tower.  It 
is  usual  practice  to  make  the  spans  between  the  towers  twice  as  long  as 
the  tower  spans,  except  where  the  height  of  the  viaduct  is  40  ft.  and 
under,  in  which  case  the  spans  are  usually  made  equal  in  length. 

From  actual  calculations  the  economic  lengths  of  spans  are  found  for 
ordinary  cases  to  be  as  follows:  30'  and  30'  for  viaducts  30  to  40  ft. 
high;  30'  and  60'  for  a  height  of  40  to  80  ft.;  and  40'  and  80'  for  a 
height  of  80  ft.  and  over. 

The  cost  of  erection  governs  the  lengths  of  spans  to  quite  an  extent. 

The  approximate  weight  of  metal  in  towers  of  ordinary  single-track 
viaducts  is  given  on  the  diagram  in  Fig.  222.  This  diagram  is  self- 
explanatory. 


296 


STRUCTURAL  ENGINEERING 


Complete  Design  of  an  Ordinary  Single-Track  Viaduct 

159.  Data.— Specifications,  A.  R.  E.  Ass'n. 

Live  Load,  Cooper's  £50. 

160.  General  Layout  of  Structure. — The  first  thing  the  designer 
needs  is  a  good  profile  of  the  crossing,  which  is  usually  furnished  by  the 
railroad  company.     Let  the  profile  shown  in  Fig.  223  be  such  a  profile  of 
the  crossing  for  which  we  are  to  design  a  viaduct. 


Cross  Frame 
Transv.  Struf 


Longitudinal  View  of  Tower 


Transverse  View  of  Benh 


Figr.  221 


Usually  (the  first  thing)  we  would  redraw  this  profile  carefully  to  a 
convenient  scale  (as  a  rule  to  a  ^  scale)  so  that  it  can  be  traced  on  the 
general  stress  sheet  for  the  structure.  Then  upon  this  profile  the  positions 
of  the  towers  are  chosen,  avoiding  the  stream  and  obtaining  as  many 
towers  and  columns  of  equal  length  as  possible,  using  the  length  of  spans 
that  is  economic  for  the  height  of  the  viaduct.  Working  in  this  manner 
we  obtain  the  general  layout  of  our  structure  as  shown  at  the  top  of  the 
general  stress  sheet,  Fig.  224.  After  this  preliminary  work  we  start  the 
detail  design  by  taking  up  the  spans  first. 

161.  Designing  of  the  Spans. — This  work  is  just  the  same  prac- 
tically as  previously  outlined  for  deck  plate  girder  bridges,  except  the 
50-ft.  spans  in  this  case  are  made  the  same  depth  as  the  60-ft.  spans  in 
order  to  simplify  construction,  the  60-ft.  spans  being  of  economic  depth. 
The  complete  stress  diagram,  or  stress  sheet,  as  we  might  say,  for  the 
three  different  lengths   of  spans   is   given   on   the  general   stress   sheet 
(Fig.  224). 

162.  Designing  of  the  Columns  and  Tower  Bracing. — In  design- 
ing a  column  the  first  thing  to  do  is  to  determine  the  maximum  concentra- 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


297 


tions  on  the  top  of  it  due  to  dead  and  live  load.  The  total  dead  load 
applied  to  the  top  of  each  column  is  equal  to  the  sum  of  the  dead-load 
reactions  9n  the  two  girders  supported  and  the  greatest  live  load  applied 

Weighiof Single  TrackV/oducf-lbwens. 

C  o  ope.r-3  E-5O  L  ooding 


130 


A.R.E.Assn.  Specs. 


4-0'  SO'          60'  70'      .     80"  90'         100' 

H&(ghrof7bwens(Frombaseofrai/fo  masonry) 
Spa 


8O'          9Of         IOO'         IIOr          I2O'        /3O'        I4O  ISO 

Height  of  Towers  (From  base  ofrai/  to  masonry) 
Fig.  222 

to  the  top  of  each  is  the  maximum  live-load  concentration  coming  from 
the  live  load  in  the  two  adjacent  spans,  just  the  same  as  in  the  case  of  an 
intermediate  floor  beam.  (Art.  148.)  These  loads  are  the  same  for  all 
columns  in  bents  1  to  9  (Fig.  224),  as  each  supports  an  end  of  a  30-  and 


uj| 

'/Grade-          >V 


Fig.  223 


GO-ft.  span.  The  load  on  the  columns  in  bent  10  is  that  coming  from  a 
30-  and  50-ft.  span  and  from  the  two  50-ft.  spans  in  the  case  of  bent  11. 
Using  the  dead-load  reactions  given  for  the  girders  on  the  stress  sheet, 
Fig.  224,  we  have 

19,000  +  6,800  =  25,800  Ibs. 

for  the  dead-load   concentration   on   all   columns   supporting  30-ft.   and 


ilil *• — Sg-s 


^Sl^®S-v 

^Ei^soisrxS  ? 


298 


SIMPLE  KAILKOAD  BEIDGES  299 

60-ft.  spans,  which  includes  the  columns  in  bents  1  to  9,  and 

6,800  +  14,400-21,200  Ibs. 
for  columns  in  bent  10  and 

14,400x2=28,800  Ibs. 

for  those  in  bent  11. 

To  obtain  the  maximum  live-load  concentration  on  a  column  we  place 
the  loading  so  that  the  heaviest  loads  (wheels)  are  near  the  column  with 
one  load  at  the  column  and  the  unit  load  in  one  adjacent  span  equal  to 

I    @5@5©     9'    ©5(7)6f®5@     8'    @    ff' 


A 


Fig.  225 

the  unit  load  in  the  other,  which  is  in  accordance  with  Art.  148  ;  the  spans 
in  this  case  being  of  unequal  length. 

Taking  first  the  columns  supporting  30-ft.  and  60-ft.  spans,  let  C 
(Fig.  225)  represent  the  column  considered.  By  placing  the  loads  as 
shown  in  Fig.  225,  we  have  (see  Table  A)  (considering  one-half  of  wheel 
12  in  each  span) 

i*M™=  2,533  Ib, 
oU 

for  the  average  unit  load  in  the  60-ft.  span,  and 

76,000 


30 


=  2,533  Ibs. 


for  the  average  unit  load  in  the  30-ft.  span. 

This  shows  that  the  criterion  is  exactly  satisfied,  and  we  need  go  no 
farther  with  the  work  of  satisfying  the  criterion.  Taking  moments  about 
A  (using  Table  A)  we  have 


for  the  reaction  on  the  column  at  C  due  to  loads  3  to  11  (inclusive),  and 
taking  moments  about  B  we  have 

=  37,300  Ibs. 


for  the  reaction  on  the  column  at  C  due  to  wheels  13  to  16  (inclusive). 
Now  adding  these  two  reactions  and  the  weight  of  wheel  12  together  we 
have 

62,000  +  37,300  +  20,000  =  119,300  Ibs. 

for  the  maximum  live-load  concentration  on  the  column  for  Cooper's  £40 
loading  and  multiplying  this  by  50/40  we  have 

119,300x55  =  149,000  Ibs. 


300  STRUCTURAL  ENGINEERING 

for  the  maximum  live-load  concentration  due  to  the  £50  loading  which 
is  the  concentration  desired. 

Then  for  the  impact  we  have 


Now  owing  to  the  columns  sloping  transversely,  known  as  the  batter, 
the  actual  stress  in  the  columns  due  to  the  above  concentrations  is  equal  to 
the  concentrations  multiplied  by  the  secant  of  the  slope  angle.  The  slope, 
or  batter,  will  be  taken  as  2  in  12,  which  is  the  usual  batter  for  such 
columns.  So  we  have  1.014  for  the  secant  of  the  slope  angle.  Then  we 
have 

25,800x1.014  =  26,000  Ibs. 

in  bents  1  to  9  (inclusive)  for  the  dead-load  stress  in  the  top  portion  of 
the  columns,  and 

149,000x1.014  =  151,000  Ibs. 
for  the  live-load  stress  in  these  columns  throughout  their  whole  length  and 

115,000x1.014  =  116,000  Ibs. 

for  the  impact. 

Now  adding  these  together  we  have 

26,000  +  151,000  +  116,000  =  293,000  Ibs. 

for  the  total  maximum  stress  in  the  top  sections  of  all  columns  in  bents 
1  to  9  (inclusive). 

We  will  next  determine  the  dead-  and  live-load  stresses  in  the 
columns  in  bent  10.  The  dead-load  concentration  on  each  of  these 


'$7)6'  (faS®     9'@5@5@.?'@    8'  ®>  8'  ©5*®  g'(?)5@ 

~^r^  -~J 


S  ©6'— _ 

30' j SO' 


SenUO 

Fig.  226 

columns  is  given  above  as  21,200  Ibs.,  and  we  will  proceed  to  determine 
the  live-load  concentration  on  them.  Placing  the  loads  as  shown  in  Fig. 
226  we  have  (considering  one-half  of  wheel  13  in  each  span) 


50 

for  the  average  unit  load  in  the  50-ft.  span,  and 

69,000 


30 


=  2,300  Ibs. 


for  the  average  unit  load  in  the  30-ft.  span. 

This  shows  that  the  load  in  the  30-ft.  span  is  a  little  too  great.     So 
let  us  place  wheel  14  at  column  C.    Then  we  have 


DESIGN  OF  SIMPLE  RAILROAD  BKIDGES  301 

for  the  average  unit  load  in  the  50-ft.  span,  and 

62,000 


30 


=  2,066  Ibs. 


for  the  average  unit  load  in  the  30-ft.  span.  So  it  is  seen  that  the 
position  of  the  wheels  shown  in  Fig.  226  will  give  the  maximum  concen- 
tration on  the  column.  So  taking  moments  about  A  (using  Table  A) 
we  have 

916 x  1.000  =30>50Qlbs       ..^::  •,,'., 

for  the  reaction  on  the  column  at  C  due  to  loads  17  to  14  (inclusive),  and 
taking  moments  about  B  we  have 

(2,036  +  102x8)1.000  ^^ 

50 

for  the  reaction  due  to  wheels  6  to  12  (inclusive). 

Now  adding  these  reactions  and  the  weight  of  wheel  13  together  we 
have 

30,500  +  57,000  +  20,000  =  107,500  Ibs. 

for  the  maximum  live-load  concentration  on  each  column  in  bent  10  due 
to  the  £40  loading.  Then  for  the  E50  loading  we  have 

55x107,500  =  134,000  Ibs. 
and  for  the  impact  we  have 

.  .•:  ;hi     134'000x(8o1Soo)=105'ooolbs-  ; 

Now  multiplying  the  dead-  and  live-load  concentrations  and  impact 
by  the  secant  of  the  slope  angle  of  the  column,  as  was  done  above,  and 
adding  all  three  of  the  results  together  we  obtain  264,000  Ibs.  for  the 


rH  .  ,  .  .-*  ts 

®S(j)  6  ®y@    8'  @  8'  ®5@5@5@    9  '  @S'®6  @5@  -S 


so  _  _  so 


Bent-// 
Fig.  227 

total  maximum  stress  in  each  of  the  columns  in  bent  10.  Next  placing 
the  wheel  loads  as  shown  in  Fig.  227  and  proceeding  in  the  same  manner 
as  above,  in  the  case  of  the  other  columns,  we  obtain 

128,600x^  =  161,000  Ibs. 


302 


STRUCTURAL  ENGINEERING 


for  the  live-load  concentration  on  each  column  in  bent  11,  and 

300      \  _ 


161,000 


121,000  Ibs. 


+  300  / 

for  the  impact  concentration. 

Then,   taking   the   dead-load    concentration    found    above    for   these 
columns,  we  have 

28,800x1.014  =  29,200  Ibs. 
for  the  dead-load  stress,  and 

161,000  x  1.014=  163,000  Ibs. 


62003^ 


Fig.  228 

for  the  live-load  stress,  and 

121,000  x  1.014  =  122,000  Ibs. 

for  the  impact,  making  in  all  314,000  Ibs.  stress  in  each  of  the  columns 
in  bent  11. 

Having  the  stresses  in  the  columns  this  far  determined  we  can  now 
write  the  dead,  live  and  impact  stress  on  them  at  the  cross-sections,  as 
shown  in  Fig.  224,  increasing  the  dead-load  stress  in  the  lower  sections 
of  the  columns  where  the  weight  of  the  part  of  the  tower  above  increases 
the  stress  an  appreciable  amount.  The  dead  weight  per  ft.  of  height  of 
towers  1-2  and  7-8  as  seen  from  the  diagram  in  Fig.  222  is  about  1,000 


DESIGN  OF  SIMPLE  KAILEOAD  BEIDGES  303 

Ibs.  So  the  weight  per  ft.  of  column  would  be  250  Ibs.  Then  the  dead- 
load  stress  in  the  lower  portion  (about  one-half  of  the  height)  of  the 
columns  in  these  towers  will  be  increased  (in  the  worst  case)  about 
250  x  28  =  7,000  Ibs.,  about  11,000  Ibs.  in  the  case  of  towers  3-4  and  5-6. 
In  tower  9-10  and  bent  11  this  increased  dead  weight  is  ignored  as  it  is 
inappreciable. 

We  shall  next  determine  the  stresses  in  the  columns  and  transverse 
bracing  due  to  wind  load.  According  to  the  specifications:  "Viaduct 
towers  shall  be  designed  for  a  force  of  50  Ibs.  per  sq.  ft.  on  one  and 
one-half  times  the  vertical  projection  of  the  structure  unloaded;  or  30  Ibs. 
per  sq.  ft.  on  the  same  surface  plus  400  Ibs.  per  linear  ft.  of  structure 
applied  7  ft.  above  the  rail  for  assumed  wind  load  on  the  train  when  the 
structure  is  fully  loaded  with  empty  cars  assumed  to  weigh  1,200  Ibs.  per 
linear  ft.  of  track." 

Let  us  first  consider  the  towers  3-4  and  5-6  (see  Fig.  224),  and  let 
the  diagram  at  (#),  Fig.  228,  represent  the  elevation  of  one  of  these 
towers  drawn  to,  say,  a  -j1^  scale.  The  wind  load  coming  onto  a  single 
bent  as  AB  would  be  that  applied  between  the  vertical  sections  SS  and 
S'S'.  This  load  will  be  considered  (in  accordance  with  practice)  to  be 
applied  at  the  joints  of  the  bents  and  on  the  girders  and  train  as  shown 
in  the  cross-section  at  (fr),  Fig.  228.  To  obtain  the  intensities  of  these 
forces  it  is  first  necessary  to  estimate  the  areas  of  the  vertical  projection 
of  the  structure  at  the  different  points.  The  area  of  the  vertical  projec- 
tion of  the  train  need  not  be  considered  as  the  load  on  it  is  given  in  the 
specification  as  400  Ibs.  per  ft.  of  track.  However,  the  train  is  usually 
considered  to  be  10  ft.  high. 

For  the  area  contributing  to  force  Fl  we  have  1.5  x  45  =  67n'  in 
round  numbers  from  ties  and  guard  rail;  3.5  x  30  =  105n'  from  the  60-ft. 
girder,  and  2.25  x  15  =  33°'  from  the  30-ft.  tower  girder,  making  in  all 
205n'  (=  67n'  +  105n/  +  33n')-  (See  Figs.  221  and  232.) 

In  the  case  of  F2  we  have  the  same  area  from  the  girders  (=  105  + 
33  =  138n/)  and  9  ft.  of  tower.  From  the  tower,  we  have  0.8  x  15  =  12n/ 
from  the  top  longitudinal  strut  (see  detail  of  tower,  Fig.  232,  also  see 
Fig.  221),  0.5x24  =  12n'  from  one-half  of  the  longitudinal  diagonal,  and 
1.5  x  9  =  14n'  from  the  column,  making  in  all  176n'  (=  138n/  +  12n/  + 
12n/  +  14°'). 

For  F3  we  have  1.5  x  18  =  27n/  from  the  column  (only). 

For  F4  we  have  1.5  x  27.7  =  42n'  from  the  column;  0.8  x  15  =  12n' 
from  the  longitudinal  strut  and  0.5  X  48  =  24n'  from  the  two  longitudinal 
diagonals  (one-half  of  each),  making  in  all  78n/  (=  42n/  +  12n'  +  24n'). 

Now  by  increasing  each  of  the  above  areas  by  one-half  (as  per 
specification)  we  obtain  the  areas  (to  the  nearest  square  feet)  indicated  at 
(6).  The  wind  load  on  the  lower  part  of  the  bent  is  transmitted  directly 
to  the  masonry  and  hence  is  not  considered. 

Taking  first  the  case  of  the  50-lb.  load  (when  the  train  is  not  on  the 
structure)  we  obtain  the  forces  indicated  at  (c)  by  multiplying  the  given 
areas  by  50.  Then  laying  off  these  forces  on  the  load  line  BC,  at  (/), 
we  obtain  the  diagram  of  the  stresses  in  the  bent,  as  shown,  by  beginning 
at  a  and  passing  around  each  joint  counter  clock- wise.  The  intensities 
of  the  stresses  thus  obtained  for  the  different  members  are  given  on  the 
bent  at  (c).  Next  for  the  case  of  the  30-lb.  load  on  the  structure  and  the 


304  STRUCTUARL  ENGINEERING 

400-lb.  per  linear  ft.  on  train,  we  have  the  forces  shown  at  (d).  Those 
applied  to  the  structure  here  are  obtained  by  multiplying  the  correspond- 
ing areas  given  at  (6)  by  30  and  the  load  on  the  train  by  multiplying 
400  by  45',  the  length  of  the  train  contributing  pressure  to  bent  AB. 
By  drawing  on  and  np  we  have  the  forces  acting  upon  a  continuous  frame 
which  can  be  graphically  analyzed  and  the  addition  of  the  members  on 
and  np  will  not  affect  the  stresses  in  the  members  of  the  bent  in  the  least. 
Then  by  laying  off  the  forces  on  the  load  line  BD  and  beginning  at  joint 
n  and  passing  around  each  joint  counter  clock- wise  we  obtain  the  diagram 
of  the  stresses  shown  at  (g).  The  intensities  of  the  stresses  thus  obtained 
for  the  different  members  are  given  on  the  bent  at  (d). 

This  work  completes  the  determination  of  the  wind  stresses  in  the 
towers  3-4  and  5-6,  and  we  shall  next  take  up  the  determination  of  the 
stresses  in  these  same  towers  due  to  the  longitudinal  force  from  the  train. 
This  force  is  known  as  traction.  It  is  the  force  exerted  along  the  rails 
when  the  brakes  are  applied  to  the  wheels  of  a  moving  train  causing  the 
wheels  to  slide  on  the  rails.  The  intensity  of  this  force,  as  is  evident,  is 


Fig.  229 


equal  to  the  coefficient  of  friction  (of  the  wheels  on  the  rails)  times  the 
vertical  load  on  the  wheels.  This  coefficient  is  designated  in  the  specifica- 
tions as  0.20.  We  will  consider  the  traction  on  each  of  the  towers,  3-4 
and  5-6,  as  coming  from  the  loads  on  the  girders  rigidly  connected  to  the 
tower  in  each  case,  or,  in  other  words,  the  loads  between  the  expansion 
points  of  the  girders.  This,  as  is  seen  from  Fig.  224  (the  points  of 
expansion  being  marked  Ex),  will  be  the  loads  on  a  30-ft.  and  60-ft.  span, 
making  in  all  90  ft.  of  continuous  load.  The  load  will  be  a  maximum 
when  the  wheels  are  in  the  position  shown  in  Fig.  229.  This  gives  a 
load  of  248,000  Ibs.  (see  Table  A)  for  Cooper's  £40  and  310,000  Ibs. 
(=  248,000  x  50/40)  for  £50.  Then  for  the  traction  force  on  each  tower 
we  have 

T  =  310,000x0.2  =  62,000  Ibs., 

which  is  applied  at  the  top  of  the  rail.  By  drawing  mk  and  Jen  at  the  top 
of  the  tower  as  shown  at  (e)  we  have  the  load  applied  to  a  continuous 
frame  which  can  be  graphically  analyzed.  The  diagram  of  stresses  in 
the  tower  shown  at  (/i),  due  to  this  62,000-lb.  force  is  obtained  by  laying 
off  AB  (representing  the  force)  as  a  load  line  and  then  beginning  at  A: 
and  passing  around  each  joint  counter  clock- wise.  The  intensities  of  the 
stresses  thus  obtained  are  given  on  the  elevation  of  the  tower  at  (<?).  The 
exact  values  of  the  stresses,  however,  are  obtained  by  multiplying  each  of 
the  given  stresses  by  the  secant  of  the  slope  angle  of  the  columns,  but  as 
each  stress  is  increased  but  a  small  amount  we  will  ignore  this  correction 
(as  is  usual  practice)  and  use  the  stresses  shown. 


DESIGN  OF  SIMPLE  EAILKOAD  BRIDGES  305 

We  now  have  all  of  the  stresses  determined  in  the  towers  3-4  and  5-6 
except  the  dead,  live,  impact,  and  traction  stresses  in  the  top  transverse 
struts  (see  Fig.  221)  of  the  bents.  These  stresses  are  due  to  the  columns 
being  inclined.  Owing  to  this  inclination  the  columns  exert  a  horizontal 
thrust  toward  each  other  upon  this  strut  causing  a  compressive  stress  in 
it  equal  to  the  vertical  load  at  the  top  of  the  column  multiplied  by  the 
tangent  of  the  slope  angle  of  the  columns.  The  slope  of  the  columns  is 
2  in  12,  so  the  tangent  of  the  slope  angle  is  0.1C66.  Then  for  the  top 
transverse  strut  we  have  the  following  stresses,  in  round  numbers: 

D=   25,800x0.1666-   4,000  Ibs.  ") 
L  =  149,000  x  0.1666  =  25,000  Ibs.    >  +  48,000  Ibs. 
7-115,000x0.1666-19,000  Ibs.   J 
T=    88,000x0.1666-15,000  Ibs. 
W=  26,000  Ibs. 

+  89,000  Ibs. 

In  order  to  design  the  column  bases  and  anchorage  it  is  necessary  to 
know  the  positive  and  negative  reactions  on  each  column.  The  maximum 
positive  reaction  on  each  is  equal  to  the  maximum  combined  stress  in  the 
bottom  section  of  the  column  divided  by  the  secant  of  the  slope  angle  of 
the  column  plus  one-fourth  of  the  weight  of  the  lower  half  of  the  tower, 
which  is  about  11,000  Ibs.  So  we  have  (see  Fig.  224) 


+R  =  11,000  +  -+567,000  Ibs. 


for  the  maximum  positive  reaction. 

The  maximum  negative  reaction,  if  there  be  such,  will  occur  when 
the  structure  is  loaded  with  a  train  of  empty  box  cars  which  is  assumed 
to  weigh  1,200  Ibs.  per  ft.  of  track,  as  per  specification. 

We  have  -98,000  Ibs.  for  the  negative  reaction  due  to  wind  load  as 
given  at  (g),  Fig.  228.  For  the  traction  force  along  each  rail  due  to  the 
1,200  Ibs.  live  load  we  have 

x  0-2  x  90  =  10,800  Ibs. 


Now  this  10,800-lb.  force  can  be  applied  to  the  top  of  the  tower  as 
was  done  with  the  62,000-lb.  force  (at  (e),  Fig.  228)  and  the  reaction 
due  to  it  determined  graphically.  But  as  the  reaction  due  to  the  62,000-lb. 
force  is  already  determined  (at  (/O),  the  one  due  to  the  10,800-lb.  force 
can  be  determined  very  readily  by  proportion  as  the  reactions  are  directly 
proportional  to  the  forces.  Hence  we  have 

x  160,000  =  -28,000  Ibs. 
for  the  negative  reaction  due  to  traction. 


306 


STRUCTURAL  ENGINEERING 


For  the  positive  reaction  due  to  the  1,200-lb.  live  load  we  have 

x  45  =  +27,000  Ibs. 


Now  adding  (algebraically)  this  and  the  dead-load  reaction,  which  is 


11,000  + 


37,000 
1.014 


-48,000  Ibs. 


(see  cross-section  of  bents,  Fig.  224),  to  the  above  negative  reactions  we 
have 

27,000  +  48,000 -  98,000  -  28,000  =  -51,000  Ibs. 

for  the  maximum  negative  reaction  that  can  occur  on  each  column  and  for 
which  anchorage  must  be  provided. 

This  method  of  determining  these 
negative  reactions  is  not  absolutely  correct 
as  the  effect  of  the  wind  and  traction  forces 
being  in  different  planes  is  not  taken  into 
account.  The  tendency  of  rotation  of  each 
tower,  as  is  obvious,  will  really  be  about 
axes  perpendicular  to  the  resultant  of  the 
two  kinds  of  forces.  However,  owing  to 
the  uncertainty  of  the  intensities  of  the 
forces  in  the  first  place  and  to  the  result- 
ing error  being  small,  the  more  exact  anal- 
ysis is  not  really  justifiable  and  hence  will 
not  be  given. 

As  we  now  have  all  of  the  stresses  determined  for  towers  3-4  and  5-6 
we  can  write  the  same  on  the  stress  sheet  (Fig.  224)  as  shown,  the  stresses 
in  the  longitudinal  bracing  on  the  elevation  of  the  towers  and  those  in  the 
transverse  bracing  and  columns  on  the  cross-section  of  the  bents,  and  we 
can  then  proceed  with  the  designing  of  the  sections  of  the  different  mem- 
bers in  these  towers. 

Taking  the  columns  first  (bents  3,  4,  5,  and  6)  let  us  assume  the 
following  section  for  the  top  portion  of  each  column : 

1— cov.  pi.  18"xf"  =  11.25n" 
2— [s  15"  x  40*       =  23.52n" 

34.77n" 

Taking  moments  about  the  center  of  the  cover  plate  (see  Fig.  230)  we  have 
-     23.52  x  7.81 


Fig.  230 


34.77 


=  5.29  ins. 


for  the  distance  to  the  center  of  gravity  of  the  assumed  section  from  the 
.center  of  the  cover  plate  (axis  x-x). 

Now  for  the  moment  of  inertia  about  axis  x-x  we  have 


7  =  347.5x2  + (2.52   x 23.52)  +  (5.29    x  11.25)  =  1,159, 
and  hence  for  the  radius  of  gyration  about  axis  x-x  we  have 


r- 


=  5.77. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  3Q7 

For  the  moment  of  inertia  about  axis  y-y  we  have 


r  =  9.39  x  2  +  (6.03    x  23.52)  +  303.75  =  1,178, 
and  for  the  radius  of  gyration  about  the  same  axis  we  have 

- 

The  distance  along  the  column  between  the  points  of  connection  of  the 
Longitudinal  bracing  is  the  greatest  unsupported  length  (as  is  seen  from 
Fig.  224)  and  hence  will  be  taken  as  the  length  of  column.  This  length 
(as  obtained  by  scale)  is  about  36'  or  432".  Then  substituting  this 
length  and  the  radius  r'  (as  the  column  would  fail  about  axis  y-y)  in  the 
column  formula,  we  have 

4.00 
p  =  16,000  -70-^-  =10,800  Ibs. 

5.8/& 

for  the  allowable  unit  stress  in  the  case  of  the  combined  dead-  and  live- 
load  and  impact  stresses.  As  the  maximum  stresses  due  to  live  load,  wind 
and  traction  are  not  likely  to  occur  simultaneously  the  unit  stress  can  be 
increased  25  per  cent  for  the  case  of  combined  dead-  and  live-load,  impact, 
wind,  and  traction  stresses,  as  per  specifications. 
So  we  have 

293,000        f 


for  the  area  required  for  the  combined  dead-  and  live-load  and  impact 
stresses,  and  for  the  combined  dead,  live,  impact,  wind,  and  traction 
stresses  we  have 

468,000  468,000 

10,800(1  +  0.25)  =T^50F  =  34.66  sq.ms., 

which  is  the  actual  required  area,  being  the  greater.     The  section  assumed 
is  satisfactory  as  it  is  about  as  near  Oe  required  area  as  we  can  get  it. 
Taking  next  the  bottom  portion  of  the  column,  let  us  assume  the 
following  section: 

1—cov.  pi.  18"xii"  =  12.37n" 
2—  [s  15"  x  50*        =  29.42n" 

41.79n// 

Taking  moments  about  the  center  of  the  cover  plate  (see  Fig.  231)  we  have 
-     29.42  x  7.84 


41.79 


=  5.52  ins. 


for  the  distance  to  the  center  of  gravity  of  the  section  from  the  center  of 
the  cover  plate. 

For  the  moment  of  inertia  about  axis  x-x  we  have 

7  -  402.7  x  2  +  (2J322  x  29.42)  +  (5^2*  x  12.37)  =  1,341, 


308 


STRUCTURAL  ENGINEERING 


and  for  the  radius  of  gyration  about  the  same  axis  we  have 


For  the  moment  of  inertia  about  axis  y-y  we  have 


/'  =  (6.05   x  29.42)  +  (11.22  x  2)  +  334.13  =  1,433, 
and  for  the  radius  of  gyration  we  have 


r  = 


The  length  of  the  bottom  portion  of  the  columns  is  about  36  ft.,  or 
432  ins.,  and  as  this  length  is  the  same  as  regards  the  two  axes,  r  the 
smaller  radius  will  be  used  in  the  column  formula.  So  we  have 


P  =  16,000  -70         =  10,680 

O.OD 

for  the  allowable  unit  stress.     Dividing  this  into  the  combined  dead,  live, 
and  impact  stress  we  have 

304,000 


10,680 


=  28.46  sq.  ins. 


for  the  required  area  of  the  column,  and  by  increasing  this  unit  stress 
25  per  cent  and  dividing  it  into  the  combined  dead,  live,  impact,  wind,  and 
traction  stress  we  have 


564,000 
13,360 


=  42.2  sq.  ins. 


Fig.  231 


for  the  required  area  which  is  the  greater 
of  the  two.  The  assumed  section  is  about 
as  near  to  the  required  section  as  is  pos- 
sible to  obtain,  so  it  will  be  used. 

We  will  next  take  up  the  designing 
of  the  transverse  bracing.  Beginning 
with  the  top  diagonals  (see  cross-section  of 
bents  3,  4,  5,  and  6,  Fig.  224)  we  have 
a  stress  of  38,000  Ibs.,  which  we  will 
assume  to  be  carried  in  tension  by  one 
system,  so  we  have 


38,000 
16,000 


=  2.37  sq.  ins. 


for  the  required  net  area  of  each  of  the  top  diagonals.  It  is  necessary  to 
use  two  angles  for  each  diagonal  in  order  to  obtain  good  details.  It  is 
seen  that  comparatively  small  angles  could  be  used  as  far  as  the  area  of 
cross-section  is  concerned,  but  as  3  J"  x  3|"  x  J"  angles  are  about  as  small 
as  is  consistent  with  good  practice  in  the  design  of  railroad  bridges  we 
will  use  2— Ls  3i"x3J"  xf"  =  4.98  -0.37  =  4.61  sq.ins.net  for  each  diag- 
onal. 


DESIGN  OF  SIMPLE  KAILEOAD  BKIDGES  309 

Now  as  these  angles  are  the  minimum  size  used  and  the  stresses  being 
less  in  the  other  transverse  diagonal  than  in  the  ones  just  considered, 
2 — Ls  3 y  x  3J"  x  f "  will  be  used  for  each  of  the  other  transverse 
diagonals. 

In  designing  the  transverse  struts  L/r  should  not  be  greater  than 
120,,  and  100  would  be  better.  The  stresses,  as  is  seen,  are  so  small  in 
these  struts  that  the  designing  of  the  sections  really  consists  in  selecting 
sections  that  will  be  sufficiently  rigid  and  provide  good  details. 

The  bottom  strut  is  about  31',  or  372",  long.     So  we  have 

373  =3.72 


100 

for  the  required  radius  of  gyration  to  give  L/r  =  100.  We  will  use 
2  —  [s  10"  x  20*  which  have  a  radius  of  3.66  and  considerable  more 
section  than  required,  but  it  is  about  as  satisfactory  a  section  as  is  obtain- 
able and  hence  will  be  used  for  the  bottom  transverse  strut. 

The  second  transverse  strut  from  the  bottom  is  about  18'-6",  or  222", 
long,  so  we  have 


for  the  required  radius.  Here  we  will  use  2  —  [s  8"  x  13.75*  which  have 
a  radius  of  2.98.  We  will  use  the  same  section  for  the  next  strut  above 
also,  in  order  to  obtain  uniform  details. 

Let  us  next  take  the  case  of  the  top  strut.  This  strut  has  a  length  of 
about  78".  Let  us  assume  2—  Ls  5"  x  3J"  x  f"  =  6.10n".  The  least 
radius  of  gyration  of  these  two  angles  taken  as  a  strut  is  1.60.  Then 
substituting  in  the  column  formula  we  have 

JVQ 

p  =  16,000  -70  -l£-  =  12,600  Ibs. 
1.60 

for  the  allowable  unit  stress  for  the  combined  dead,  live,  and  impact 
stresses.  So  we  have 

48,000 


for  the  required  area  of  cross-section,  and  increasing  the  above  unit  stress 
25  per  cent  and  dividing  it  into  the  combined  dead,  live,  impact,  wind,  and 
traction  stresses  we  have 

89,000 


for  the  required  area  which  is  the  greater.     So  our  assumed  section  is 
about  correct  and  hence  will  be  used. 

We  will  next  take  up  the  designing  of  the  longitudinal  bracing  in 
towers  3-4  and  5-6.  Here  each  diagonal  will  be  designed  to  carry  the 
96,000-lb.  stress  in  tension,  assuming  only  one  system  to  act  at  a  time. 
So  we  have 

96,000  ,     ^ 

68q.ms.  (net) 


DESIGN  OF  SIMPLE  EAILBOAD  BKIDGES  3H 

for  the  required  net  area  of  cross-section  of  each  diagonal.  Then  for  each 
diagonal  we  can  use  (counting  out  of  each  angle  one  rivet  hole)  2  —  Ls 
6"  x  4"  x  f"  =  7.22  -  0.75  =  6.47  sq.  ins.  (net)  . 

We  will  next  consider  the  longitudinal  struts,  which  are  really 
columns  30'  or  360"  long.  The  maximum  stress  of  62,000*  occurs  on 
the  intermediate  one.  Let  us  assume  a  section  composed  of  2  —  [s  10"x 
20*  =  11.76°".  Then  we  have 


r     3.66 
and  we  also  have 

p  =  16,000  -  70  —  =  9,110  Ibs. 
o.oo 

for  the  allowable  unit  stress.    Dividing  this  into  the  stress  we  have 

62,000 


for  the  required  area  which  is  much  less  than  the  area  of  the  assumed 
section,  yet  we  will  use  the  assumed  section  for  each  of  the  longitudinal 
struts  as  the  L/r  is  about  correct.  This  completes  the  design  for  the 
towers  3-4  and  5-6,  and  the  sections  can  be  written  on  the  stress  sheet  as 
shown  (Fig.  224). 

The  other  towers  can  be  designed  in  the  same  manner  as  shown  above 
for  towers  3-4  and  5-6  and  then  the  stress  sheet  can  be  finished  as  shown 
in  Fig.  224. 

163.  Detail  Drawings. — After  the  stress  sheet  (Fig.  224)  is 
completed  a  general  drawing  of  one  (at  least)  of  the  towers  should  be 
made  in  order  to  show  the  kind  of  details  desired  unless  standard  draw- 
ings for  such  details,  previously  made,  are  available.  The  tower  selected 
for  detailing  should  be  one  of  medium  height  so  as  to  obtain  average 
details.  In  this  case  we  will  select  tower  1-2  (see  stress  sheet,  Fig.  224). 
The  general  details  for  the  tower  will  be  sufficiently  shown  by  drawing 
the  details  of  one  column  and  the  principal  details  of  the  bracing  connect- 
ing to  the  column.  The  first  thing  to  do  is  to  make  large  scale  detail 
sketches  of  the  column  cap,  base,  and  of  the  principal  intermediate  points. 
From  these  sketches  we  determine  the  exact  slopes  and  clearances  for  the 
bracing,  in  fact  all  essential  details  are  worked  out  on  these  sketches  and 
simply  transferred  to  the  finished  general  drawing.  Working  in  this 
manner  we  obtain  the  general  drawing  for  the  tower  1-2  shown  in 
Fig.  232. 

The  calculations  for  these  details  are  quite  simple.  The  cap  is  made 
as  small  as  is  consistent  with  good  details.  The  I-beam  diaphragm  at 
the  top  of  the  column  is  for  the  purpose  of  distributing  the  loads  from  the 
girders  equally  to  the  two  channels.  There  should  be  a  sufficient  number 
of  rivets  in  each  side  of  this  diaphragm  to  transmit  one-half  of  the 
maximum  end  shear  on  the  longer  girder  minus  the  dead-load  end  shear 
on  the  shorter  girder.  So  in  this  case  we  have  (see  stress  sheet,  Fig.  224) 

243_,000 
2 


312  STRUCTURAL  ENGINEERING 

for  the  shear  that  these  rivets  are  to  transmit.  Dividing  this  by  7,200 
(the  value  of  a  J"  shop  rivet  in  single  shear)  we  obtain  practically  16 
rivets.  Sixteen  is  the  number  used. 

The  bolts  connecting  the  girders  to  the  columns  should  be  sufficiently 
large  to  transmit  the  shear  due  to  traction,  neglecting  the  friction  on  the 
expansion  end.  Taking  the  case  of  the  60-ft.  span,  the  heaviest  load  will 
occur  when  wheels  2  to  11  (inclusive)  are  on  the  span.  This  (for  Cooper's 
E5Q,  see  Fig.  151)  gives  a  load  of  205,000  Ibs.  per  girder. 

Then,  we  have  0.2  x  205,000  =  41,000  Ibs.  for  the  horizontal  thrust 
on  each  girder  which  must  be  taken  by  the  bolts  at  the  fixed  end  of  the 
girder.  We  have  four  bolts  taking  this,  so  each  must  take  one-fourth  of 
it  or  10,250  Ibs.  This  calls  for  1-J"  bolts,  as  the  area  of  cross-section  of 
each  is  0.994n",  and  stressing  them  10,000  Ibs.  per  square  inch  gives 
9,940  Ibs.,  which  is  about  the  required  value.  The  same  size  bolts  are 
used  for  the  tower  spans  in  order  to  have  uniform  sizes,  although  the 
shear  due  to  traction  in  that  case  does  not  require  them  to  be  so  large. 

The  column  base  should  have  sufficient  area  so  as  not  to  stress  the 
masonry  more  than  600  Ibs.  per  square  inch  (see  specifications).  The 
maximum  reaction  on  bents  1  and  2  (see  stress  sheet,  Fig.  224)  is  512,000 
Ibs.  Dividing  this  by  600  we  obtain  853  sq.  ins.  for  the  required  area  of 
bearing  on  the  masonry.  The  masonry  plate  or  base  plate  used  has 
(30"  x  29")  870  sq.  ins.,  which  is  about  the  correct  area.  The  masonry 
plate  should  be  symmetrically  placed  in  reference  to  the  center  of  gravity 
of  the  column  and  the  details  should  be  such  that  the  pressure  from  the 
column  is  quite  uniformly  distributed  over  the  plate. 

The  anchor  bolts  should  be  large  enough  to  take  the  maximum  nega- 
tive reaction  (uplift)  on  the  column  at  16,000  Ibs.  per  square  inch.  The 
splice  in  the  column  is  considered  to  be  a  butt  joint,  that  is,  the  top  section 
of  the  column  is  considered  to  bear  firmly  against  the  bottom  section  so 
that  the  stress  is  transmitted  from  one  to  the  other  without  the  aid  of  the 
rivets  in  the  splice.  In  that  case  there  need  be  only  enough  rivets  in  the 
splice  to  hold  the  column  in  line.  Just  how  many  to  use  in  such  cases 
must  be  determined  by  mere  judgment. 

The  calculation  for  the  other  details  is  mostly  a  matter  of  develop- 
ing the  sections  in  the  bracing  which  the  student  should  have  no  trouble 
in  doing.  Take,  for  example,  the  longitudinal  diagonals.  Each  of  these 
diagonals  is  composed  of  2— Ls  6"  x  4"  x  f  "  =  7.22  -  0.75  =  6.47  sq.  ins.  (net) . 
Multiplying  this  net  area  (of  the  two  angles)  by  16,000  Ibs.  we  have 

6.47x16,000  =  103,500  Ibs. 

for  the  tensile  strength  of  each  diagonal.  Dividing  this  by  6,000  Ibs.  we 
obtain  about  17 — -J"  field  rivets  to  develop  the  section.  Eight  on  a  side 
are  used  (in  the  end  connections),  which  is  about  correct. 

Take  next  the  longitudinal  struts  which  are  composed  of  2 — [s 
10"x20#  =  11.76n".  The  strength  of  these  struts  as  given  in  the  last 
Article  is  9,110  Ibs.  per  square  inch.  Then  for  the  total  strength  of  each 
we  have 

11.76x9,110  =  107,000  Ibs. 


DESIGN  OF  SIMPLE  KAILKOAD  BKIDGES  313 

Dividing  this  by  6,000  Ibs.  we  obtain  about  18 — J"  field  rivets.  Nine  on 
a  side  are  used  (in  the  end  connections),  which  is  correct.  The  riveting 
in  the  end  connections  and  splices  of  the  transverse  bracing  is  obtained  in 
the  same  manner. 

The  stress  sheet  (Fig.  224)  and  the  general  drawing  of  the  tower  1-2 
(Fig.  232)  are  quite  sufficient  for  general  drawings  in  this  case  as  the 
work  is  quite  similar  throughout.  However,  there  should  always  be  a 
sufficient  number  of  general  drawings  to  fully  show  the  character  of  the 
work.  All  special  towers  should  really  be  detailed. 

From  these  general  drawings  the  bridge  company,  obtaining  the 
contract  for  fabricating  the  structure,  works  up  the  shop  drawings  for 
the  work. 

In  working  up  the  shop  drawings  usually  the  first  thing,  after  large 
scale  sketches  of  the  principal  details  of  the  towers  are  made,  is  to  make 
a  drawing  showing  the  location  of  the  anchor  bolts.  This  is  known  as  the 
masonry  plan.  This  drawing  is  used  by  the  party  building  the  sub- 
structure and  is  usually  the  first  drawing  called  for.  The  material  as  a 
rule  is  next  ordered  and  line  drawings  of  the  towers  are  made  upon  which 
the  lengths  of  all  members  are  given,  which  are  usually  computed  by  the 
aid  of  logarithmic  tables.  After  this  the  making  of  the  shop  drawings 
proper  (and  bills)  proceeds.  The  shop  drawings  for  the  girders  are  prac- 
tically the  same  as  previously  shown  for  deck-plate  girder  bridges.  The 
shop  drawings  for  the  towers  are  quite  easy  to  make ;  the  columns  should 
be  on  separate  sheets  from  the  bracing.  All  should  be  drawn  as  much  in 
their  relative  position  as  possible  to  aid  in  checking  and  also  in  identifying 
the  members. 

We  will  next  make  a  preliminary  estimate  of  the  weight  and  cost  of 
metal  in  the  structure. 

164.  Preliminary  Estimate  of  Weight  of  Metal  and  Cost. — 
Taking  up  the  weight  of  the  girders  first  we  have  the  following: 

Weight  of  metal  in  one  60-ft.  span. 
Weight  of  one  girder. 

1— web  84"  x  f"  x  107.1*  x  60' 6,426  Ibs. 

4_Ls  6"  x  6"  x  I"  x  24.2*  x  60' 5,808  Ibs. 

1— cov.  pi.  14"  x  £ "  x  29.76*  x  60' 1,786  Ibs. 

l_cov.  pi.  14"  x  f "  x  29.76*  x  44' 1,309  Ibs. 

2— cov.  pis.  14"  x  i"  x  23.8*  x  29' 1,380  Ibs. 

26— stiff.  Ls  5"  x  3i"  x  f "  x  10.4*  x  7' 1,893  Ibs. 

8— end  stiff.  Ls  5"  x  3J"  x  TV  x  12.0*  x  7' 672  Ibs, 

4— fillers  7"  x  f"  x  14.88*  x  6' 357  Ibs. 

8— sp.  pis.  10"  x  f "  x  21.25*  x  2.4' 408  Ibs. 

4— sp.  pis.  14"  x  f"  x  29.76*  x  4.3' 512  Ibs. 

2_sole  pis.  12"  x  f "  x  30.6*  x  1.2' 74  Ibs. 

6— fillers  3J"  x  f"  x  7.44*  x  6' 268  Ibs. 

20,893  Ibs. 
rivets,  3%    (=20,890  x  0.03) 627  Ibs. 


Total  weight  of  1  girder 21,520  Ibs. 


Total  weight  of  2  girders. 43,040  Ibs. 


314  STRUCTURAL  ENGINEERING 

Weight  of  one  frame. 

2— Ls  3i"  x  3J"  x  f "  x  8.5*  x  6.1' 104  Ibs. 

2— Ls  3£"  x  3|"  x  f"  x  8.5*  x  8.2' 140  Ibs. 

4— pis.  12J"  x  f  "  x  15.94*  x  1.1' 70  Ibs. 

1— pi.  9"  x  f"  x  11.48*  x  0.8' 9  Ibs. 

323  Ibs. 
rivets   22  Ibs. 

Total  weight  of  1  frame 345  Ibs. 

5 

Total  weight  of  5  frames 1,725  Ibs. 

Weight  of  laterals  and  lateral  plates. 

8— Ls  3J"  x  3£"  x  f"  x  8.5*  x  9' 612  Ibs. 

7— pis.  12"  x  f"  x  15.3*  x  2.8' 300  Ibs. 

15— pis.  12"  x  |"  x  15.3*  x  1.0' 230  Ibs. 

2— pis.  12"  x  f"  x  15.3*  x  2.0' 61  Ibs. 

rivets 27  Ibs. 

Total  weight  of  laterals  and  lateral  plates 1,230  Ibs. 

Summary  of  weight. 

2— girders    43,040  Ibs. 

5— frames 1,720  Ibs. 

laterals  and  plates 1,230  Ibs. 

Total  weight  of  metal  in  1— 60-ft.  span 45,990  Ibs. 

Weight  of  metal  in  one  30-ft.  span. 
Weight  of  one  girder. 

1— web  54"  x  f"  x  68.85*  x  25' 1,721  Ibs. 

4— Ls  6"  x  6"  x  f"  x  24.2*  x  30' 2,904  Ibs. 

2— end  pis.  30"  x  f"  x  38.25*  x  6' 459  Ibs. 

4— sp.  pis.  12"  x  f"  x  25.5*  x  3.5' 357  Ibs. 

4— Ls  6"  x  6"  x  £"  x  19.6*  x  4'  (bent) 314  Ibs. 

4— Ls  6"  x  6"  x  |"  x  19.6*  x  7' 549  Ibs. 

4— fillers  6"  x  f "  x  12.7*  x  6' 305  Ibs. 

14— Ls  5"  x  3£"  x  f"  x  10.4*  x  4.5' 655  Ibs. 

2— pis.  12"  x  f"  x  25.5*  x  2' 102  Ibs. 

2— sole  pis.  12"  x  |"  x  30.6*  x  1.2' . 74  Ibs. 

7,440  Ibs. 
rivets,   2% 150  Ibs. 

Total  weight  of  1  girder 7,590  Ibs. 

2 

Total  weight  of  2  girders 15,180  Ibs. 

Weight  of  frames. 

2 — end  frames  =  345  x  2 690  Ibs. (from 60- ft. span) 

1 — interior  frame 320  Ibs.  (computed) 

Total  weight  of  3  frames., 1,010  Ibs. 


DESIGN  OF  SIMPLE  KAILKOAD  BRIDGES  315 

1  230 
Weight  of  laterals  and  plates  -br~x  30  =  615*  (from  60-ft.  span) 

Summary  of  weight. 

2— girders 15,180  Ibs. 

3— frames 1,010  Ibs. 

laterals  and  plates 615  Ibs. 

Total  weight  of  metal  in  1 — 30'  tower  span 16,805  Ibs. 

Weight  of  metal  in  one  50-ft.  span. 

1— web  84"  x  f"  x  107.1*  x  50' 5,355  Ibs. 

4_Ls  6"  x  6"  x  J"  x  33.1*  x  50' 6,620  Ibs. 

22— -Ls  5"  x  34"  x  f"  x  10.4*  x  7' 1,602  Ibs. 

4— Ls  5"  x  3J"  x  Ty  x  12.0*  x  7' 336  Ibs. 

4— fillers  7"  x  £"  x  20.83*  x  6' 500  Ibs. 

4— fillers  31"  x  J"  x  10.42*  x  6' 250  Ibs. 

splice  plates  (see  60-ft.  span) 650  Ibs. 

sole  plates  (see  60-ft.  span) 74  Ibs. 

15,387  Ibs. 
rivets,    3% 450  Ibs. 

Total  weight  of  1  girder 15,837  Ibs. 

O 


Total  weight  of  2  girders 31,664  Ibs. 

Lateral  system  =  (about  same  as  50-ft.  span,  Art.  137) .  .    2,776  Ibs. 
Total  weight  of  metal  in  1 — 50-ft.  span 34,440  Ibs. 

Weight  of  metal  in  tower  1-2. 

(Detailed,  Fig.  232.) 

One  Column. 

2— [s  15"  x  33*  x  23.75' 1,568  Ibs. 

1—cov.  pi.  18"  x  f"  x  22,95*  x  23.7' 544  Ibs. 

2_[s  15"  x  50*  x  26.6' 2,660  Ibs. 

l_cov.  pi.  is"  x  •&"  x  26.78*  x  26.6' 712  Ibs. 

Total  weight  of  main  section 5,484  Ibs. 

Cap. 

1— cap  plate  20J"  x  f"  x  52.28*  x  3.1' 162  Ibs. 

1__I  10"  x  30*  x  2.3' 69  Ibs. 

2— fills.  5"  x  i"  x  4.25*  x  2.3' 20  Ibs. 

2_i_s  6"  x  4"  x  f"  x  20*  x  1.7' 68  Ibs. 

2_Ls  6"  x  4"  x  f"  x  20*  x  1.0' 40  Ibs. 

*1— gus.  pi.  29"  x  f"  x  36.97*  x  2.4' 89  Ibs. 

1— gus.  pi.  36"  x'f"  x  45.90*  x  2.6' 120  Ibs. 

2_Ls  3£"  x  3-|"  x  f "  x  8.5*  x  1.4' 24  Ibs. 

2— pis.  16"  x  f"  x  20.4*  x  1.75' 71  Ibs. 

2_[_s  5"  x  3J"  x  |"  x  10.4*  x  1.0' 21  Ibs. 

684  Ibs. 


*  Average  width  and  length  of  plates  are  given. 


316  STRUCTURAL  ENGINEERING 

Intermediate  details. 

3— tie  pis.  18"  x  f  "  x  22.95*  x  1.5' 103  Ibs. 

2— lat.  pis.  14"  x  f"  x  17.85*  x  3' 107  Ibs. 

4_sp.  pig.  12"  x  f  "  x  15.3*  x  1.85' 113  Ibs. 

2— sp.  pis.  18"  x  f"  x  22.95*  x  2.0' 92  Ibs. 

2— lat.  pis.  16"  x  f"  x  20.4*  x  3.0' 122  Ibs. 

2— lat.  pis.  20"  x  f"  x  25.5*  x  3.4' 173  Ibs. 

128  ft.  of  2J"  x  f"  latt.  bars  @  3.19*  per  ft 408  Ibs. 

1,118  Ibs. 
Base. 

l_base  plate  30"  x  f"  x  89.25*  x  2.5' 223  Ibs. 

2_Ls  6"  x  6"  x  f"  x  24.2*  x  1.1' 53  Ibs. 

2_Ls  6"  x  4"  x  f"  x  20*  x  1.6' 64  Ibs. 

2_Ls  6"  x  6"  x  f"  x  24.2*  x  0.4' 19  Ibs. 

2_Ls  6"  x  6"  x  f"  x  24.2*  x  0.8' 39  Ibs. 

4— Ls  5"  x  31"  x  f"  x  10.4*  x  1.0' 42  Ibs. 

2— Ls  5"  x  31"  x  10.4*  x  2.0' 42  Ibs. 

1_L  6"  x  6"  x  f  "  x  24.2*  x  2.4' 58  Ibs. 

2— fills.  W  x  f "  x  7.44*  x  1.5' 22  Ibs. 

l_pl.  9"  x  |"  x  H.48*  x  1.5' 17  Ibs. 

2— Ls  31"  x  31"  x  f"  x  8.5*  x  1.5' 26  Ibs. 

1— gus.  pi.  30"  x  f"  x  38.25*  x  3.3' 126  Ibs. 

1— gus.  pi.  28"  x  f "  x  35.7*  x  1.8' 64  Ibs. 

2— gus.  pis.  30"  x  f"  x  38.25*  x  1.75' . 134  Ibs. 

929  Ibs. 

Summary  of  weight  of  details. 

Cap 684  Ibs. 

Int.  details 1,118  Ibs. 

Base 929  Ibs. 

Rivets 200  Ibs. 

2,931  Ibs. 


2  931 
Percentage  of  details  =  S  ,*  —  53%  of  main  section. 


Summary  of  weight  of  one  column  (bents  1  and  3). 

Details    2,931  Ibs. 

Main  section 5,484  Ibs. 

8,415  Ibs.  =  total  weight  of  one  column. 
4 


33,660  Ibs.  =  total  weight  of  four   columns. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  317 

Longitudinal  bracing. 
Longitudinal  strut. 

2_[s  10"  x  20*  x  28.4' 1,136  Ibs. 

4— tie  pis.  12"  x  f "  x  15.3*  x  1.25' 77  Ibs. 

164  ft.  latt.  bars  of  21"  x  f"  @  2.87*  per  ft 470  Ibs. 

rivets 53  Ibs. 

(Details  -  44%.)  1,736  Ibs. 

6_ 

Weight  of  metal  in  six  struts 10,416  Ibs. 

Diagonal. 

2— Ls  6"  x  4"  x  f"  x  12.3*  x  35.5'.  . 873  Ibs. 

2— tie  pis.  12"  x  f"  x  15.3*  x  1.25' 38  Ibs. 

48  ft.  latt.  bars  2J"  x  f"  @  2.87*  per  ft 138  Ibs. 

rivets 20_  Ibs. 

(Details  =  22%.)  1,069  Ibs. 

4 

Weight  of  metal  in  four  diagonals 4,276  Ibs. 

Diagonal  (spliced). 

4— Ls  6"  x  4"  x  f"  x  12.3*  x  17.5; 861  Ibs. 

4— tie  pis.  12"  x  f"  x  15.3*  x  1.25' 77  Ibs. 

2— sp.  pis.  13"  x  |"  x  16.57*  x  3.5' 116  Ibs. 

44  ft.  latt.  bars  2J"  x  f"  @  2.87*  per  ft 126  Ibs, 

rivets 24  Ibs. 

(Details  =  40%.)  1,204  Ibs. 

4 

Weight  of  metal  in  four  diagonals 4,816  Ibs. 

Total  weight  of  longitudinal  bracing 19,508  Ibs. 

Transverse  bracing  (bents  1  and  2). 
Strut  (beginning  at  top  of  bent). 

2_Ls  5"  x  3J"  x  f  "  x  10.4*  x  4.0' 83  Ibs. 

7  ft.  21"  x  f"  lat.  bars  @  2.87* 20  Ibs. 

rivets 2  Ibs.     105  Ibs. 

(Details,  27%.) 
Diagonal. 

2_Ls  3J"  x  3J"  x  f"  x  8.5*  x  12.75' 217  Ibs. 

2— tie  pis.  9"  x  f"  x  11.48*  x  1.0' 23  Ibs. 

16  ft.  lat.  bars  21"  x  f"  @  2.87* 46  Ibs. 

rivets 11  Ibs.     297  Ibs. 

(Details,  36%.) 


318  STRUCTURAL  ENGINEERING 

Diagonal. 

2—Ls  34"  x  34"  x  f"  x  8.5*  x  3.75' 64  Ibs. 

2— Ls  34"  x  3i"  x  f"  x  8.5*  x  8.4' 143  Ibs. 

4— tie  pis.  9"  x  f"  x  11.48*  x  1.0' 46  Ibs. 

2— sp.  pis.  12"  x  f"  x  15.3*  x  3.0' 92  Ibs. 

14  ft.  latt.  bars  2J"  x  f"  @  2.87* 40  Ibs. 

rivets 16  Ibs.     401  Ibs. 


(Details,  93%.) 
Strut. 

2— [s  8"  x  13.75*  x  8.75' 240  Ibs. 

4— tie  pis.  12"  x  f "  x  15.3*  x  0.8' 49  Ibs. 

20  ft.  lat.  bars  2J"  x  f"  @  2.87* 57  Ibs. 

rivets ,  12  Ibs.     358  Ibs. 


(Details,  49%.) 
Diagonal. 

2— Ls  3J"  x  34"  x  f"  x  8.5*  x  17.2' 292  Ibs. 

2— tie  pis.  9"  x  f"  x  11.48*  x  1.0' 23  Ibs. 

34  ft.  lat.  bars  2J"  x  f"  @  2.87* 97  Ibs. 

rivets 16  Ibs.     428  Ibs. 


(Details,  46%.) 
Diagonal. 

2— Ls  34"  x  34"  x  f"  x  8.5*  x  6.6' 112  Ibs. 

2— Ls  34"  x  34"  x  f"  x  8.5*  x  10.0' 170  Ibs. 

4— tie  pis.  9"  x  f"  x  11.48*  x  1.0' .  : 46  Ibs. 

30  ft.  lat.  bars  2J"  x  f"  @  2.87* 86  Ibs. 

2— sp.  pis.  12"  x  f"  x  15.3*  x  3.0' 92  Ibs. 

rivets  .  20  Ibs.     526  Ibs. 


(Details,  87%.) 
Strut. 

2— [s  8"  x  13.75*  x  13.3' 366  Ibs. 

4— tie  pis.  12"  x  f  "  x  15.3*  x  0.8' 49  Ibs. 

40  ft.  lat.  bars  2J"  x  f"  @  2.87* 115  Ibs. 

rivets  .  40  Ibs.     570  Ibs. 


(Details,  56%.) 
Diagonal. 

2— Ls  34"  x  34"  x  f"  x  8.5*  x  27.2' 462  Ibs. 

2— tie  pis.  9"  x  f"  x  11.48*  x  1.0' 23  Ibs. 

48  ft.  lat.  bars  2J"  x  f"  @  2.87* 137  Ibs. 

rivets 26  Ibs.     648  Ibs. 

(Details,  40%.) 
Diagonal. 

2—\_s  34"  x  34"  x  f"  x  8.5*  x  10.6' 180  Ibs. 

2— Ls  34"  x  34"  x  f"  x  8.5*  x  16.0' 272  Ibs. 

4— tie  pis.  9"  x  f"  x  11.48*  x  1.0' 46  Ibs. 

44  ft.  lat.  bars  2J"  x  f"  @  2.87* 126  Ibs. 

2— sp.  pis.  12"  x  f"  x  15.3*  x  3.0' 92  Ibs. 

rivets  .  30  Ibs.     746  Ibs. 


(Details,  65%.) 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES  319 

Strut. 

2—  [s  10"  x  20#  x  20.5' 820  Ibs. 

4— tie  pis.  12"  x  |"  x  15.3*  x  0.8' 49  Ibs. 

72  ft.  lat.  bars  2±"  x  f  "  @  2.87* 206  Ibs. 

rivets 50  Ibs.  1,125  lbs; 

(Details,  37%.) 

Weight  of  transverse  bracing  in  one  bent  =  5,204  Ibs. 

Weight  of  transverse  bracing  in  two  bents  or  tower  =  10,408  Ibs. 

Summary  of  weight  of  metal  in  tower  1-2. 

4  columns 33,660  Ibs. 

Long,  bracing 19,508  Ibs. 

Trans,  bracing 10,408  Ibs. 

Total 63,676  Ibs. 

Proceeding  in  the  same  manner  the   following  weights  are  found: 

Weight  of  metal  in  each  of  the  towers  3-4  and  5-6. 

4  columns  (36%  details) 52,200  Ibs. 

Long,  bracing ,  21,820  Ibs. 

Trans,  bracing 14,440  Ibs. 

Total 88,460  Ibs. 

Weight  of  metal  in  tower  7-8 64,900  Ibs. 

Weight  of  metal  in  tower  9-10 42,380  Ibs. 

Weight  of  metal  in  bent  11 8,420  Ibs. 

Summary  of  weight  of  metal  in  structure. 

5— 60-ft.  spans  @  45,990  Ibs 229,950  lbs-1 

5_30-ft.  spans  @  16,805  Ibs 84,025  Ibs.  j>  382,855  Ibs. 

2— 50-ft.  spans  @  34,440  Ibs 68,880  Ibs.J 

l_tower   (1-2) 63,676  Ibs.  1 

2— towers  (3-4  and  5-6)   @  88,460  Ibs 176,920  Ibs.  j 

l_tower   (7-8) 64,900  Ibs.  j>  356,296  Ibs. 

1— tower   (9-10) 42,380  Ibs. 

1— bent    (10) 8,420  Ibs.J 

Total  weight  of  metal 739,151  Ibs. 

Cost  of  structure   (superstructure). 

Girders,  382,855  Ibs.  @  3^ $11,486 

Towers,  356,296  Ibs.  @  3^ 12,470 

Total  cost  of  steel  work  (except  anchor  bolts) $23,956 

This  price  is  only  a  fair  average  pound  price. 

165.  Double-Track  Viaducts. — The  ordinary  double-track  viaduct 
has  a  double  line  of  spans,  that  is,  four  lines  of  track  girders,  which,  as 
a  rule,  are  supported  upon  cross-girders  as  shown  in  Fig.  233  instead  of 


320 


STRUCTURAL  ENGINEERING 


resting  directly  upon  the  top  of  columns.  The  towers,  as  for  general 
design,  are  practically  the  same  as  for  single-track  viaducts.  The  concen- 
trations at  the  top  of  the  towers  due  to  dead  and  live  load  and  impact  are 
just  twice  as  much  as  for  single-track  viaducts  and  consequently  the 
stresses  in  the  columns  due  to  the  same  will  be  twice  as  much,  provided 
of  course,  the  columns  in  the  two  cases  have  the  same  batter.  We  assume 
that  two  trains  move  abreast  over  the  double-track  structure.  The  trac- 
tion is  just  twice  as  much  for  a  double-track  as  it  is  for  a  single-track 
viaduct,  while  the  wind  pressure  is  practically  the  same  for  the  two  struc- 
tures. (See  specifications  for  wind  load.) 

The  method  of  procedure  in  designing  double-track  viaducts  is  the 
same  as  for  single-track  viaducts  except  the  transverse  bracing  is  sub- 


Fig.  233 

jected  to  live-load  stress  and  impact  when  only  one  track  is  loaded.  This 
is  due  to  the  unequal  thrusts  exerted  by  the  two  columns  upon  the  top 
strut,  or  cross-girder.  As  an  illustration,  suppose  the  right-hand  track  of 
a  double-track  viaduct  (see  Fig.  233)  to  be  loaded  with  the  maximum  live 
load  and  no  live  load  on  the  left-hand  track.  The  right-hand  column  will 
receive  most  of  the  load,  as  is  evident.  Let  V  —  the  concentration  on  the 
right-hand  column  and  V  the  concentration  on  the  left-hand  column  and 
let  <f>  represent  the  slope  angle  of  the  columns.  The  horizontal  thrust  on 
the  cross-girder  exerted  by  the  right-hand  column  is  equal  to  Ftan<£,  and 
that  exerted  by  the  left-hand  column  is  equal  to  F'tan<£.  Now  if  these 
two  thrusts  were  equal  they  would  just  balance  each  other  and  there 
would  be  no  live-load  stress  in  the  transverse  bracing.  But  as  they  are 
unequal  the  transverse  bracing  must  transmit  the  difference  (which  pro- 
duces a  horizontal  shear)  down  to  the  masonry.  This  force  (=  Ftan<f>  - 
F'tanc^)  can  be  assumed  as  applied  at  the  bottom  flange  (considered  as  a 
strut)  of  the  cross-girder  (just  exactly  as  a  wind  load)  and  the  stresses 
in  the  transverse  bracing  due  to  it  graphically  determined. 


DESIGN  OF  SIMPLE  RAILEOAD  BEIDGES 


321 


The  maximum  stress  in  the  columns  occurs  when  the  two  tracks  are 
fully  loaded,  at  which  time  no  live-load  stress  occurs  in  the  transverse 
bracing. 

166.    Analytical  Method   of  Determining   Stresses   in   Tower 
Bracing.  —  Let  Fig.   234  represent  a  transverse   view   of  a  bent  of  an 
ordinary   viaduct,   acted   upon   by    forces   as   indi- 
cated.     Let   us   first   suppose   that   the   horizontal 
wind  forces  F,  Fl,  F2,  and  F3  alone  are  acting 
and  that  the  stress  in  the  diagonals  AD,  CN,  and 
in  the  strut  CD,  due  to  these  forces,  is  to  be  deter- 
mined. 

To  determine  the  stress  in  the  diagonal  AD,< 
first  assume  the  part  of  the  bent  (forces  and  all) 
below  the  section  mm  removed.  Then  the  part 
above  this  section  mm  is  held  in  equilibrium  by 
the  forces  F,  Fl,  F2,  S,  S3,  and  £4,  where  S,  S3, 
and  S4:  represent  the  stress  in  members  AD,  AC, 
and  BD,  respectively.  The  stress  S  is  what  we 
desire.  By  prolonging  the  lines  of  action  of 
the  forces  (stresses)  S3  and  $4  and  taking  mo- 
ments about  their  point  of  intersection,  0,  we 
eliminate  these  forces  from  the  equation  of  moments 
and  we  have 


Fig.  234 


from  which  we  obtain  (in  known  quantities) 

cF  +  hFl  +  JcF2 

O  =  -  :  - 


Similarly,  to  determine  the  stress  S2  in  diagonal  CN,  assume  the 
part  of  the  bent  below  the  section  nn  removed  and  taking  moments  about 
0  we  have 

cF  +  hFl  +  kF2  +  rF3  +  dS2  =  0, 

from  which  we  obtain  the  required  stress 
cF  +  hFl  +  JcF2  +  rF3 


To  determine  the  stress  /SI  in  the  strut  CD  assume  the  part  of  the 
bent  below  the  section  oo  removed  and  then  taking  moments  about  0  we 
have 

cF  +  hFl  +  kF2  +  rF3  +  rSl  =  0, 

from  which  we  obtain  the  required  stress 

cF  +  hFl  +  kF2  +  rF3 

Sl  = • 

r 

Instead  of  wind  loads,  as  just  considered,  suppose  there  be  an  eccen- 
trically applied  load  W  acting  upon  the  bent.  Then  taking  moments 
about  0,  as  before,  we  have 


322 


DESIGN  OF  SIMPLE  KAILEOAD  BEIDGES 

from  which  we  obtain 

We 


323 


and  similarly  we  obtain 


We 


We 

?-?- 

d 


Webs- 


Fig    236 


Eccentrically   applied   loads,   as   just   considered,   occur   mostly   on 

viaducts  built  on  curve  and  on  double-track  viaducts  when  only  one  track 

is  loaded  with  live  load. 

If  desired,  the  stresses  in  the  columns  can  be  determined  by  taking 

moments  about  panel  points.  For  example,  the  stress  S±  in  BD  can  be 
obtained  by  passing  the  section  mm  and  taking 
moments  about  A.  By  passing  section  oo  and 
taking  moments  about  D  the  stress  S3  in  AC  can 
be  determined,  and,  by  taking  moments  at  the  same 
time  about  C  the  stress  S6  in  DN  can  be  determined. 
Similarly,  by  passing  the  section  nn  and  taking 
moments  about  N  the  stress  S5  in  CM  can  be 
determined. 

The  only  objection  to  the  above  method  is  that 

the  length  of  some  of  the  lever  arms  is  troublesome  to  determine  unless 

it  be  determined  by  scale,  which,  as  a  rule,  is  not  a  very  satisfactory 

manner  unless  special  care  be  taken. 

167.      General    Remarks.  —  Viaduct   columns   are   sometimes   built 

without  cover  plates  as  shown  in  Fig.  235.     There  is  no  question  but  that 

a  column  with  a  cover  plate  is  better  than  one  without.     Whenever  a 

greater  area  is  required  than  can  be  obtained  by  using  two  channels  and  a 

cover  plate  a  section  like  the  one  shown  in  Fig.   236  should  be  used. 

However,  in  case  there  be  just  a  few  column  sections  in  a  structure 

requiring  more  area  than  is  contained  in  two 

channels  and  a  cover  plate,  the  area  may  be 

increased  so  that  that  type  of  section  can  be 

used  by  riveting  a  plate  to  the  back  of  each 

channel. 

The   batter    of   the    columns    in    single- 

track  viaducts  varies  from  1  j  in  12  to  3  in  12, 

for  very  low  bents.      Two  in   12   is   a  very 

common  batter.     The  batter  of  the  columns  in 

double-track  viaducts,  as  a  rule,  is  less  than 

in  the  case  of  single-track  structures.     The 

batter  should  be  sufficient  to  limit  the  negative 

reactions  at  the  bases  of  the  columns  to  a 

reasonable    intensity,    at    least.      In    fact    it 

would  be  better  to  have  no  negative  reaction 

at  all,  but  as  a  rule  it  is  not  practical  to  have  that  condition  in  the  case  of 

single-track  viaducts. 

There  should  be  sufficient  material  in  each  pedestal  (which  is  usually 

made  of   concrete)    to   properly   distribute  the   positive   reaction   of  the 

column  supported  and  at  the  same  time  have  sufficient  weight  to  resist 


X"C*  Anchor  both 


Fig.  237 


324 


STRUCTURAL  ENGINEERING 


the  negative  reaction  of  the  column.     The  anchor  bolts  should  extend  well 
down  into  the  pedestals  as  indicated  in  Fig.  237. 

In  the  case  of  viaduct  towers  having  no  horizontal  struts,  as  shown 
in  Fig.  221,  the  two  systems  of  bracing  are  usually  considered  to  act 
simultaneously,  which  requires  that  each  member  of  the  bracing  be 
designed  for  compression  as  well  as  for  tension,  but  at  the  same  time, 
however,  the  stress  in  each  is  only  half  as  much  as  obtains  in  the  type 
of  bracing  shown  in  Fig.  220,  assuming  only  one  system  to  act  at  a  time. 

DRAWING  ROOM  EXERCISE  NO.  6 

Design  and  make  stress  sheet  of  a  single-track  viaduct  for  the 
following  crossing: 

Elevation  of 
Elevation  of  Ground 
640 
588 
579 
568 


Base  of  Rail 

648 
648 
648 
648 


Creek 


Station 
400 

400  +  81 
401 

401  +  70 

402  +  13  564 
402  +  20  554 
402  +  40  554 

402  +  49  566 

403  567 

403  +  44  564 
404+  5  608 

404  +  41  613 
404  +  80  626 
405+  2  646 

Data: 

Live  load,  Cooper's  jE50  loading. 

Specifications,  A.  R.  E.  Ass'n. 

Dead  load,  to  be  determined  by  student. 

TRUSS  BRIDGES 

168.  Preliminary. — Truss  bridges  are  used  for  spans  100  ft.  in 
length  and  over.  With  regard  to  the  location  of  the  track  there  are  two 
types  of  truss  bridges,  the  through  and  the  deck  bridge,  the  same  as  in 


648 
648 
648 
648 
648 
648 
648 
648 
648 
648 


Wit  Truss  (b) 

Flff.  288 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


325 


the  case  of  plate  girders.   The  tnrough  type  is  used  wherever  the  under 
clearance  will  not  permit  of  the  use  of  the  deck  type. 

The  four  trusses  shown  in  Fig.  238  are  the  most  common  types  used 
for  through  bridges.  Pratt  trusses  are  used  for  spans  up  to  175  ft.  in 
length  and  as  a  rule  are  riveted  trusses.  Curved  chord  Pratt  trusses  are 
used  for  spans  from  200  ft.  to  325  ft.  long  and  as  a  rule  are  pin-connected 


Fig.  240 


trusses.  Pettit  trusses,  both  (a)  and  (b)  types,  are  used  for  long  spans, 
350  ft.  in  length  and  over. 

There  are  other  types  of  bridge  trusses  which  are  used  to  some 
extent.  These  will  be  considered  later. 

In  the  case  of  deck  bridges,  the  Pratt  truss  is  practically  always 


326  STRUCTURAL  ENGINEERING 

used.     The  end  panels,  however,  are  modified  either  as  shown  at  (a)  or 
(6),  Fig.  239. 

The  diagram,  Fig.  240,  gives  the  names  of  the  different  parts  of  an 
ordinary  Pratt  truss  bridge,  but  the  same  names  hold  for  bridge  trusses 
:n  general. 

Complete  Design  of  a  150-Ft.  Single-Track  Through  Riveted 
Pratt  Truss  Span 

169.  Data.— 

Length  =  6  panels  @  25'-0"  =  150'-0"  c.c.  end  pins. 
Width  =  16'-0"  c.c.  of  trusses. 
Height  =  30'-0"  c.c.  chords. 
Stringers  spaced,  6'-6"  c.c. 
Live  Load,  Cooper's  .E50  loading. 
Specifications,  A.  R.  E.  Ass'n. 

170.  Design  of  25-Ft.   Stringers.  —  (For  detail   of   stringers,   see 
Fig.   290.)      For  dead  load  on  stringers   we  have,  from   (1)    Art.   124, 
B>=12x25  +  100  +  400  =  8001bs.  per  ft.  of  span  or  400  Ibs.  per  ft.  of 
stringer.     Then,  using  this  load,  we  have 

252  x  12  =  375,000  inch  Ibs. 


12.5' 


for  the  maximum  bending  moment  due  to  dead  load. 

It  is  seen  from  Table  A  that  the  maximum  live-load  moment  will 

likely  occur  when  wheels  2  to  5  are  on  the  stringer,  and,  according  to  Art. 

88,  they  will  be  in  the  position  shown  in  Fig.  241. 

Taking  moments  about  A  (Fig.  241),  to 
find  the  reaction  R  (using  Table  A),  we  find 
first  the  moment  of  wheels  3,  4  and  5  about  2 

*L  W*  WfflW  J  W |B  by    passing   down    the    vertical    line    through 

wheel  2  to  the  zig-zag  line,  then  to  the  right 
to  the  vertical  line  through  wheel  5  and  the 
figure  600,  just  to  the  right  of  this  line,  is  the 

moment  of  wheels  3,  4  and  5  about  2,  in  thousands  of  foot  pounds. 

Then  multiplying  the  total  weight,  in  thousands  of  pounds,  of  the 

wheels,  2  to  5,  by  3.75  (see  Art.  46)  and  adding  this  product  to  the  600, 

we  have  the  moment  of  the  wheels,  2  to  5,  about  A.     So  we  can  write 

the  equation  of  moment  about  A  as 

600  +  (80  x  3.75)  -  (R  x  25)  =0, 
from  which  we  obtain 


25' 


/CO 

for  the  reaction  at  B. 

Then  taking  moments  about  4,  as  the  maximum  moment  occurs  under 
that  wheel  (see  Art.  88),  we  have  the  moment  of  R  about  wheel  4  minus 
the  moment  of  wheel  5  about  wheel  4  for  the  maximum  bending  moment; 
that  is,  we  have 

M"  =  36,000  x  11.25  -  5  x  20,000  =  305,000  ft.  Ibs., 

for  the  maximum  live-load  bending  moment  due  to  Cooper's  £40  loading, 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  327 

and  multiplying  this  by  50/40  and  by  12  we  have  4,575,000"*  for  the 
desired  maximum  live-load  moment  in  inch  pounds,  provided  the  correct 
group  of  wheels  were  selected,  which  can  be  ascertained  by  trial,  in  case 
of  doubt. 

Then  for  the  impact  moment  we  have 


7  =  4,575,000  x  -— =  4,223,000  inch  Ibs. 
325 

Now  adding  the  above  moments  together  we  have 

M  =  375,000  +  4,575,000  +  4,223,000  =  9,173,000  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  stringer. 

Next  let  us  assume  the  web  as  f"  thick.     Then  substituting  in  (5) 
of  Art.  113  (stringers,  as  a  rule,  never  have  cover  plates),  we  have 

ff=l.$ 

for  the  economic  depth.     So  we  will  use  a  web  48' 


25'  V  For  the  dead-load   end  shear  on  the  stringer. 

Fig  242  WG   haVC 

400x12.5  =  5,000  Ibs., 

and  placing  the  wheels  as  shown  in  Fig.  242  and  taking  moments  about 
the  support  B  (using  Table  A)  we  have 


for  the  maximum  end  shear  due  to  Cooper's  £40  and 

|5  x  56,500  =  70,650  Ibs.,  say  71,000  Ibs., 
due  to  Cooper's  E5Q  loading,  and  for  impact  we  have 
71,000  x  |55  =  66,000  Ibs.  (about) 


and  adding  we  have 

5,000  +  71,000  +  66,000  =  142,000  Ibs. 

for  the  total  maximum  end  shear  on  the  stringer. 

Now  for  the  unit  shear  on  the  assumed  web  we  have 

142,000 


and  substituting  this  in  (1)  of  Art.  118,  we  have 

d=  1_  (12,000  -  7,888)  =  39  ins.  (about) 

for  the  maximum  distance  allowed  between  the  stiffeners  near  the  ends 
of  the  stringer,  and,  as  this  is  more  than  half  the  depth  of  the  girder 
and  as  the  minimum  thickness  of  web  allowed  by  the  specifications  is 
f  in.,  we  will  use  a  48"  x  f"  web. 


328  STRUCTURAL  ENGINEERING 

To  determine  the  area  of  flanges  the  first  thing  to  do  is  to  approxi- 
mate the  effective  depth  of  the  stringer.  Assuming  that  6"  x  6"  angles 
will  be  required  we  find  from  Table  6,  or  from  some  handbook,  that  the 
average  distance  from  the  back  to  the  center  of  gravity  of  these  angles 
is  about  1.75  ins.,  and  taking  the  depth  of  the  girder  as  J"  more  than 
the  web,  we  have 

48.25  -  1.75  x  2  =  44.75,  say  45  ins., 

for  the  approximate  effective  depth.      Dividing  this  into   the  maximum 
total  bending  moment,  we  have 

=  204,000  lb,  (about), 


for  the  stress  in  each  flange  (see  Art.  112),  and  dividing  this  by  16,000 
we  have 

204,000 

=  12.75sq.  ins., 


16,000 
for  the  net  area  of  each  flange.     Let  us  try 

2— Ls  6"  x  6"  x  \"  =  (5.75  x  2)  -  1  =  10.50n"  net 
$  area  of  web  =    2.25n"net 

12.75°"  net 

This  gives  exactly  the  area  indicated.     Now  checking  back,  we  have 
48.25 -1.68x2  =  44.89  ins. 

for  the  actual  effective  depth,  which  differs  so  little  from  the  assumed 
that  the  area  of  the  flanges  would  be  changed  but  very  little,  and  hence 
recalculation  of  flanges  is  unnecessary. 

The  stiffeners  on  stringers  are  usually  made  of  3J"  x  3 J"  angles. 
There  is  no  rational  way  of  determining  them.  Owing  to  the  planing  off 
of  the  ends  of  the  stringers  to  obtain  exact  length,  the  end  stiffeners  are, 
as  a  rule,  made  ^$"  to  -J"  thicker  than  the  intermediate  stiffeners,  which 
are  of  minimum  (f")  thickness. 

The  laterals — bracing  for  stringers — are  designed  the  same  as  for 
any  deck  plate  girder.  (See  Art.  135.) 

171.  Design  of  Intermediate  Floor  Beams. — (For  details  of 
floor  beams,  see  Fig.  291.)  The  dead  load  on  an  intermediate  floor  beam 
consists  of  the  dead-load  concentrations  from  the  stringers  and  the  weight 
of  the  floor  beam  itself. 

Each  concentration  from  the  stringers  is  equal  to  twice  the  end  shear 
on  one  stringer  and  such  floor  beams  weigh,  depending  on  details,  from 
3,000*  to  3,800*  each.  So  let  us,  in  this  case,  assume  3,600*  as  the 
weight  of  one  intermediate  floor  beam  complete,  or  225*  per  ft.  of  floor 
beam.  Then,  for  dead  load  we  have  the  case  shown  in  Fig.  243,  and 
for  the  maximum  bending  moment  due  to  dead  load,  we  have 

10,000  x  4.75  x  12  =  570,000  in.  Ibs. 
£x225xl62x!2   =    86,000  in.  Ibs. 

656,000  in.  Ibs. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  329 

and  for  the  maximum  end  shear,  due  to  the  same  load,  we  have 

10,000  =  11,800  Ibs. 

To  determine  the  maximum  live-load  bending  moment  and  end  shear 
on  the  floor  beam  we  must  first  satisfy  the  criterion  of  Art.  148  for 
maximum  live-load  concentration  on  the  floor  beam.  That  is,  the  loads  in 
the  two  adjacent  panels  must  be  as  nearly  equal  as  possible  and  at  the 
same  time  the  heaviest  loads  must  be  near  the  floor  beam,  with  one  load 
exactly  at  it. 


Fig.  243  Fig.  244 

By  trial  (using  Table  A)  it  will  be  found  that  the  maximum  con- 
centration on  the  floor  beams  will  occur  when  the  wheels  are  in  the 
position  shown  in  Fig.  244,  as  the  criterion  is  most  nearly  satisfied  with 
them  in  that  position,  that  is,  the  heaviest  wheels  are  near  the  floor  beam, 
wheel  13  at  it,  and  the  load  in  panel  BC,  consisting  of  wheels  14,  15,  and 
16  (wheel  17  is  exactly  over  the  floor  beam  at  C  and  hence  is  not  in  the 
panel  #C),  weighs  46,000  Ibs.,  while  the  load  in  panel  AB,  consisting  of 
wheels  10,  11,  and  12,  weighs  50,000  Ibs.  This  is  as  close  as  the  crite- 
rion can  be  satisfied,  as  will  be  found  by  further  investigation.  So, 
taking  moments  (using  Table  A)  about  A  to  find  the  concentration  at  B 
due  to  wheels  10,  11,  and  12,  we  have 


r  =  (420  +  50  x  7)        r  =  30>800  lbs'> 

ivO 

and  taking  moments  about  C  to  find  the  concentration  at  B  due  to  wheels 
14,  15,  and  16,  we  have 


r'  =         x  1,000  =  24,850  Ibs. 
&  o 

Now  adding  these  concentrations  and  the  weight  of  wheel  13  together, 
we  have 

R  =  r  +  r'  +  20,000  =  30,800  +  24,850  +  20,000  =  75,650  Ibs., 

for  the  maximum  live-load  concentration  due  to  Cooper's  £40  loading, 
and  multiplying  this  by  50/40  we  have  94,562*,  say  94,500#,  for  the 
maximum  live-load  concentration  due  to  Cooper's  -E50  loading. 

Then  for  the  live  load  on  each  intermediate  floor  beam  we  have  the 
case  shown  in  Fig.  245,  and  for  the  maximum  live-load  moment,  taking 
moment  about  either  C  or  D,  we  have 

4.75  x  94,500  x  12  =  5,386,000  inch  Ibs., 
and  for  the  impact  we  have 


5,386,000  x         =  4,616,800,  say,  4,620,000  inch  Ibs. 


330 


STRUCTURAL  ENGINEERING 


(The  load  extends  over  two  panel  lengths,  or  50  ft.) 

Now  adding  the  above  dead,  live,  and  impact  moments,  we  have 
M  =  656,000  +  5,386,000  +  4,620,000  =  10,662,000  inch  Ibs., 

for  the  total  maximum  bending  moment  on  the  floor  beam,  and,  as  is 
seen  from  Fig.  245,  the  maximum  live-load  shear  is  94,500*  and  the 
impact  =  94,500x300/350  =  81,000*.  So  by  adding  these  to  the  above 
dead-load  shear  we  have  11,800  +  94,500  +  81,000  =  187,300*  for  the  total 
maximum  end  shear  on  the  floor  beam. 

The  depth  of  the  floor  beam  is  governed  practically  by  the  depth 
of  the  stringers.  The  bottom  laterals  and  the  bottom  flanges  of  the  floor 
beams  should  be  in  the  same  plane,  as  the  bottom  flanges  of  the  beams 
act  as  struts  in  the  bottom  lateral  system,  and  there  should  be  sufficient 
distance  from  the  bottom  of  the  stringers  down  to  the  bottom  of  the  floor 
beam  to  permit  the  laterals  to  pass  beneath  the  stringers,  which  usually 
requires  from  4"  to  6",  depending  on  the  size  of  the  laterals,  and  the  dis- 
tance from  the  top  of  the  floor  beams  down  to  the  top  of  the  stringers 
is  usually  about  3".  In  this  case  we  will  assume  3"  as  the  distance 
from  the  top  of  the  floor  beams  down  to  the  top  of  the  stringers  and 
5"  from  the  bottom  of  the  stringers  down  to  the  bottom  of  the  floor 
beams,  which  gives  56J"  for  the  depth  of  the  floor  beam  as  shown  in 
Fig.  246.  It  will  be  found  upon  investigation  that  floor  beams,  as  a  rule, 
are  deeper  than  the  economic  depth. 


1 


e'-e 


Floor  Beam. 


Fig.  245 


Fig.  246 


After  having  decided  upon  the  depth  of  the  floor  beam,  the  next 
thing  is  to  approximate  the  effective  depth.  Assuming  that  each  flange 
of  the  floor  beam  will  be  composed  of  2 — 6"  x  6"  angles,  we  can  take 
about  the  average  distance  from  the  back  of  the  angles  to  their  centers 
of  gravity,  which  is  given  in  Table  6  as,  say,  1.75".  Then,  for  the 
approximate  effective  depth,  we  have 

56.25 -(1.75x2)  =52.75,  say  53  ins. 
Dividing  this  into  the  above  maximum  bending  moment,  we  have 

10,662,000  +  53  =  201,000  Ibs.  (about) 

for  the  flange  stress,  and  dividing  this  by  16,000  we  have 
201,000  -5- 16,000  =  12.56  sq.  in. 

for  the  required  net  area  of  each  flange. 

Assuming  a  56"  x  TV'  weD  which  has  a  section  of  24.5n"  (=  56  x  •&), 

let  us  try 

2— Ls  6"  x  6"  x  £"  =  (5.75  x  2)  -  1  =  10.50*"  net 
-J  area  of  cross-section  of  web  =    3.06n"  net 


13.56n"  net 


DESIGN  OF  SIMPLE  EAILKOAD  BEIDGES  331 

This  gives  ln"  more  than  is  required.  2 — Ls  6"  x  6"  x  Ty  have  a 
section  more  nearly  equal  to  the  required  area,  but  the  specifications 
require  the  angles  to  be  -J"  thick  if  the  outstanding  leg  is  6",  so,  to  com- 
ply with  the  specifications,  the  above  section  will  be  used. 

It  will  be  seen  that  the  true  effective  depth  here  is  slightly  less  than 
the  assumed,  as  the  distance  from  the  back  of  the  6"  x  6"  x  \"  angles  is 
1.68".  So,  for  the  actual  effective  depth,  we  have 

56.25 -(1.68x2) -52.89  ins. 

This,  however,  is  so  near  the  assumed  effective  depth  that  recalculations 
are  unnecessary. 

For  the  maximum  unit  shear  on  the  floor  beam  we  have 

1 87,300  -f  24.5  =  7,644  Ibs.  per  sq.  in. 
Now  substituting  this  value  for  s  in  (1),  Art.  118,  we  have 

d  =  15  (12,000  -  7,644)  =  47.6  ins., 

for  the  allowable  distance  between  stiffeners.  The  shear  is  practically 
zero  between  the  stringers,  so  that  no  stiffeners  are  required  there,  and 
the  clear  distance  from  stringer  to  truss  is  4/-9'/  minus  about  7"  for 
half  width  of  truss  and  6"  (at  least)  for  connections,  leaving  3'  -  8",  or 
44",  for  the  actual  value  of  d,  while  47.6"  is  allowed,  so  the  assumed 
web  is  quite  thick  enough.  In  fact  it  could  be  thinner,  but  the  rivets, 
spaced  along  the  flanges  between  the  stringers  and  trusses,  would  be 
closer  than  desired,  if  a  thinner  web  be  used,  and  hence  the  assumed 
56"  x  T7<r"  web  will  be  used,  even  though  it  is  a  little  thicker  than 
required. 

The  end  stiffeners  on  the  floor  beams  are  usually  6"  x  6"  x  J"  angles, 
placed  upon  fillers  as  shown  in  Fig.  291. 

172.  Design  of  End  Floor  Beams. — The  dead  load  on  an  end 
floor  beam  consists  of  the  two  dead-load  concentrations  from  the  stringers, 


,50  f*r  ft. 


A— 1^^^ 
I      *g     1 


4.'-9"  I    6-  6' 


4-9' 


I6-O" 


Fig.  247  Fig.  248 

each  of  which  is  equal  to  the  dead-load  end  shear  on  one  stringer,  and  the 
weight  of  the  floor  beam  itself.  The  dead-load  end  shear  on  a  stringer,  as 
given  in  Art.  170,  is  5,000*,  and  such  end  floor  beams  weigh  about  2,400#, 
or,  say,  150*  per  foot.  Then  the  dead  weight  on  each  end  floor  beam  is 
as  shown  in  Fig.  247. 

For  the  maximum  bending  moment,  due  to  this  dead  load,  we  have 

5,000x4.75x12   =285,000  inch  Ibs. 

|  x  150  x  16~2  x  12  =    57,600  inch  Ibs. 

342,600,  say  343,000  inch  Ibs., 


332  STRUCTURAL  ENGINEERING 

and  for  the  maximum  end  shear,  due  to  the  same  dead  load,  we  have 
5,000  +  150  x  8-6,200  Ibs. 

The  maximum  live-load  concentrations  from  the  stringers  on  an  end 
floor  beam  are  each  equal  to  the  maximum  live-load  end  shear  on  one 
stringer,  which  is  given  in  Art.  170  as  71,000*.  So  the  live  load  on  an 
end  floor  beam  is  as  shown  in  Fig.  248. 

For  maximum  bending  moment,  due  to  this  live  load,  we  have 

71,000  x  4.75  x  12  =  4,047,000  inch  Ibs., 
and  for  the  impact  we  have 

4,047,000  X  |22  =  3,735,000  inch  Ibs., 


and  for  maximum  end  shear,  due  to  live  load,  we  have  71,000,  and  for 
end  shear,  due  to  impact,  we  have 


71,000  x         =  66,000  Ibs.  (about). 
o#5 

Now  adding  the  above  moments,  we  have 

343,000  +  4,047,000  +  3,735,000  =  8,125,000  inch  Ibs., 

for  the  total  maximum  bending  moment,  and  adding  together  the  above 
end  shears  we  have 

6,200  +  71,000  +  66,000  =  143,200  Ibs., 

for  the  total  maximum  end  shear  on  an  end  floor  beam. 

Now  assuming  53"  for  the  effective  depth,  the  same  as  for  the  inter- 
mediate floor  beam,  and  dividing  it  into  the  above  moment,  we  have 

=  153,000  Ibs. 


for  the  flange  stress,  and  dividing  this  by  16,000,  we  have 

153,000 


16,000 


=  9.56sq.  in., 


for  the  required  net  area  of  each  flange.  Now  it  is  readily  seen,  from 
Table  6,  that  6"  x  6"  angles  are  too  large  for  these  flanges  and  hence  we 
will  use  6"  x  4"  angles  with  the  6"  leg  along  the  web  to  provide  for  the 
flange  rivets. 

Let  us  try,  assuming  a  56"  x  f "  web  (same  depth  as  the  intermediate 
floor  beams), 

2_Ls  6"x4"xTy'=:4.18x2-0.87  =  7.49°" net 
£  area  of  web  =  (56  x  J)  -r  8      =  2.62n"net 

10.11n"net 

This  flange  section  is  about  correct.  It  is  0.55n"  greater  than  called  for 
above,  but  the  assumed  effective  depth  is  greater  than  the  actual,  which  is 
really  56.25  -  (2  x  1.96)  =  52,33". 


DESIGN  OF  SIMPLE  EAILROAD  BEIDGES  333 

Dividing  the  moment  by  this,  we  have 

8,125,000  -r  52.33  =  155,000, 
and  dividing  this  by  16,000*,  we  have 

155,000 


16,000 


=  9.68  sq.  ins. 


for  the  actual  flange  area  required,  which  is  0.43n"  less  than  the  above 
assumed  section,  but  the  above  section  will  be  used,  as  6"  x  4"  x  f  "  angles 
are  too  small,  giving  a  net  area  of  0.59n"  less  than  required. 

Dividing  the  end   shear  by  the   area  of  cross-section   of  the   web, 
we  have 

=  6,819  Ibs., 


21 

for  the  actual  unit  shear  on  the  web.  Substituting  this  value  for  s  in  (1), 
Art.  118,  we  have 

4=sjr  (12,000  -6,819)  =48.5  ins., 

for  the  allowable  distance  between  stiffeners,  but  as  the  unsupported  dis- 
tance between  stringers  and  truss  is  about  3'  -  8",  or  44",  according  to 
the  preceding  article,  no  stiffeners  are  needed  and  the  assumed  56"  xf" 
web  will  be  used.  The  actual  allowable  stress  on  it  is  determined  by  sub- 
stituting 44"  for  d  in  (1),  Art.  118,  and  solving  for  s.  Thus  we  have 

44  =1  (18,000  -.), 

from  which  we  obtain 

*  =  7,306  Ibs., 

for  the  actual  allowable  unit  stress.  Now  dividing  this  into  the  maximum 
end  shear,  we  have 

143,200 

=  19.57  sq.ms., 


for  the  required  area  of  cross-section  of  the  web.  The  area  of  the 
56"  x  f"  web,  which  we  propose  to  use,  is  21n",  which,  as  is  seen,  is 
more  than  is  required,  but  the  area  of  a  56"x^V  web,  which  is  the  next 
in  size,  contains  only  17.5n".  Therefore,  a  thinner  web  could  not  be 
used  even  if  the  specifications  would  permit  it.  From  this  it  is  seen  that 
the  assumed  56"  x  f  "  web  is  as  near  the  correct  section  as  is  possible 
to  obtain  and  hence  will  be  used. 

This  completes  the  design  of  the  floor  system,  and  next  we  will  take 
up  the  design  of  the  trusses. 

173.  Determination  of  Dead-Load  Stresses  in  Trusses.—  The 
dead  load  of  an  ordinary  bridge  is  considered  as  uniformly  distributed 
along  the  span  but  applied  to  the  trusses  only  at  the  panel  points  or 
joints,  one-  third  at  the  top  joints  and  two-thirds  at  the  bottom  joints,  in 
the  case  of  through  bridges,  and  just  the  reverse  in  the  case  of  deck  spans. 


334 


STRUCTURAL  ENGINEERING 


This  load  consists  of  the  weight  of  metal  in  the  span  and  the  weight 
of  the  deck. 

From  (4),  Art.  124,  we  have 

p  =  7x150 +  660  =  1,710  Ibs., 

for  the  approximate  weight  of  the  metal  per  foot  of  span,  and  adding 
400*  for  the  weight  of  the  deck,  we  have 

1,710 +  400  =  2,110  Ibs., 

for  the  total  assumed  dead  load  per  foot  of  span,  or  2,110/2  =  1,055*  per 
foot  of  truss. 

Multiplying  this  by  25,  the  panel  length  in  feet,  we  have 

W=  1,055  x  25  =  26,375  Ibs., 

for  the  panel  load  on  each  truss.  One-third  of  this  is  considered  as 
applied  at  each  top  joint  and  two-thirds  at  each  bottom  joint.  However, 
this  assumption  affects  only  the  verticals,  and,  as  far  as  the  stresses  in 
the  other  members  are  concerned,  the  full  panel  load  could  be  considered 
at  the  bottom  joints,  or  top  either,  as  for  that. 

After  having  thus  computed  the  panel  load,  the  next  thing  to  do  is 
to  draw  a  sketch  of  the  truss,  as  shown  in  Fig.  249  (as  a  guide),  and  then 
compute  the  values  of  tan  6  and  sec  6. 

9  f\ 

Tan  6  =  ^:=  0.8333  and  sec  0=1.299,  say  1.3. 
oU 

In  determining  the  dead-load  stresses  in  the  web  members  (diagonals 
and  verticals),  we  can  begin  either  at  the  center  of  the  span  or  at  the 
end.  The  dead-load  stresses  in  these  members  can  really  be  determined 
by  mere  inspection. 

c/=25-0°, 


Fig.  249 

Beginning  at  the  center  of  the  span  and  considering  the  post  dD,  it 
is  obvious  that  this  post  can  do  nothing  more  than  carry  the  load  applied 
to  the  top  of  it,  which  is  one-third  of  the  panel  load;  so,  evidently,  the 
stress  in  it  is  26,375  -=-3  =  8,791*,  say  9,000*,  compression. 

The  one-third  of  a  panel  load  transmitted  by  the  post  dD  down  to 
joint  D  combines  with  the  two-thirds  at  D  and  undoubtedly  one-half  of 
this  combination  is  transmitted  to  the  two  ends  of  the  truss.  Therefore, 
the  shear  in  panel  CD  is  one-half  of  a  panel  load  and  the  stress  in  the 
diagonal  cD  is 


sec  6  = 


x  1.3  =  17,143,  say  17,000  Ibs.  tension. 


DESIGN  OF  SIMPLE  EAILKOAD  BEIDGES  335 

The  stress  in  the  post  cC  is  equal  to  the  half  panel  load  coming  from  panel 
point  D,  or  the  vertical  pull  of  diagonal  cD,  plus  the  one-third  panel 
load  at  c.  So,  for  the  stress  in  post  cC,  we  have 


^=        26,375  =  21,980,  say  22,000  Ibs. 

fy  O  O  O 

Likewise,  it  is  evident  that  the  diagonal  bC  must  transmit,  so  to 
speak,  the  \  W  coming  from  the  central  panel  point  D,  the  J  W  from  c, 
and  the  f  W  from  C,  making  1J  W  in  all,  which  is  the  shear  in  panel  BC 
and  must,  undoubtedly,  be  equal  to  the  vertical  component  of  the  stress  in 
the  diagonal  bC,  and  hence  the  stress  in  the  diagonal  is  equal  to  IJx 
26,375  x  sec  0  =  51,430#,  say  51,000*,  tension. 

The  member  bB,  known  as  a  hanger  or  hip  vertical,  as  is  readily 
seen,  can  carry  nothing  more  than  the  load  hung  on  (so  to  speak)  at  the 
joint  B,  and  hence  the  stress  in  it  is  f  x  26,375  =  17,582#,  say  18,000#, 
tension. 

It  is  evident  that  the  end  post  bA  must  transmit  -J  W  from  the  cen- 
tral panel  point  D,  the  J  W  from  c,  the  f  W  from  C,  the  f  W  from  B, 
and  the  J  W  from  b,  making  in  all  2J  W,  which  is  the  shear  in  panel  AB 
and  equal  to  the  vertical  component  of  the  stress  in  this  member  bA.  So 
the  stress  in  it  is  equal  to  2J  x  26,375  x  sec  0  =  85,718#,  say  85,000#,  com- 
pression. 

Now,  if  we  begin  at  the  end  of  the  span  instead  of  at  the  center,  as 
we  did  above,  we  have  first  the  reaction  R  =  2%  W  at  A.  (This  is  obtained 
from  mere  observation.)  As  this  reaction  acts  vertically  the  end  post 
bA  must  resist  it,  as  the  end  post  is  the  only  member  at  point  A  having  a 
vertical  component.  So  we  have  2-J  W  sec  0  for  the  stress  in  the  end  post, 
the  same  as  found  above  for  this  member.  The  stress  in  the  hanger  bB 
is  f  W,  as  found  before.  The  shear  in  panel  BC  is  equal  to  the  reaction 
R  (=2J  W)  minus  the  $  W  at  b  and  f  W  at  B,  so  that  we  have  2J  W- 
rr=lj  W  for  the  shear  in  panel  BC.  This,  evidently,  must  be  equal  to 
the  vertical  component  of  the  stress  in  diagonal  bC,  as  that  member  is  the 
only  member  in  that  panel  having  a  vertical  component  force,  and  hence 
the  stress  in  diagonal  bC  is  equal  to  1^  W  sec  6,  the  same  as  found  before. 
Likewise,  the  stress  in  diagonal  cD  is  equal  to  the  shear  in  panel  CD 
times  sec#.  The  shear  is  equal  to  R-2W  =  $W,  and  hence  the  stress  is 
equal  to  -J  W  sec  6,  the  same  as  found  before.  The  stress  in  the  post  cC 
is  equal  to  the  vertical  component  of  the  diagonal  cD,  which  is  J  W  (the 
shear  in  panel  CD)  plus  the  J  W  at  c,  making  f  W,  the  same  as  found 
before,  or  it  is  equal  to  the  vertical  component  of  diagonal  bC,  which  is 
11  W  (the  shear  in  panel  BC),  minus  the  f  W  at  C. 

The  expression  for  the  stresses  in  the  web  members  can  be  written 
on  the  members,  as  shown  on  the  right  half  of  the  truss  (Fig.  249),  and 
the  numerical  results  of  the  same  may  be  found  very  quickly  by  the  use 
of  the  slide  rule. 

For  the  chord  stresses  we  can  assume  the  full  panel  loads  as  applied 
at  the  bottom  panel  points,  in  order  to  simplify  the  work. 

Let  us  first  write  out  the  formulas  for  the  chord  stresses.  For  the 
stress  S  in  bottom  chords  AB  and  BC,  we  have,  taking  moments  about 
point  b  (see  Fig.  249), 


336  STRUCTURAL  ENGINEERING 

Transposing,  we  have 

S  =  2±  JF|  =  (2J)  Wtan6  .............................  (1). 

For  the  stress  S2)  hi  chord  be,  we  have,,  taking  moments  about  C, 


from  which  we  obtain 

S2  =  5W^-W^=5Wtan6-Wtan6  =  4:Wtaii  6  ..........  (2) 

for  the  stress  in  chord  be.     We  obtain  the  same  thing  for  bottom  chord 
CD  by  taking  moments  about  c. 

For  the  stress  S3,  in  top  chord  cd,  we  have,  taking  moments  about  D, 


from  which  we  obtain 


We  thus  have  an  expression  for  the  stress  in  each  chord  member. 
Substituting  the  numerical  values  in  (1),  we  have 

S  =  2$x  26,375x0.8333  =  54,945,  say  55,000  Ibs. 

tension  for  the  stress  in  chords  AB  and  EC,  and  by  substituting,  like- 
wise, the  numerical  values  in  (2)  and  (3)  we  obtain  87,672*,  say 
88,000#,  for  the  stress  in  the  chord  members  be  and  CD,  and  98,631*, 
say  99,000*,  for  the  stress  in  the  chord  member  cd.  This  completes  the 
calculations  for  the  dead-load  stresses  in  the  trusses,  as  the  structure  is 
symmetrical  about  the  center  of  the  span. 


Fig.  250 

It  will  be  seen  from  Formulas  (1),  (2),  and  (3)  that  WtanQ  is  a 
constant,  and  after  the  coefficients  are  written  out  the  stresses  in  the 
chords  could  be  quickly  determined  by  the  use  of  the  slide  rule. 

These  coefficients  can  be  written  down  on  the  chord  members  by 
mere  inspection.  For  example,  let  Fig.  250  represent  a  truss  of  nine 
equal  panels. 

Taking  moments  about  b  for  the  stress  in  the  bottom  chords  AB  and 
BCf  we  have  4Wd/h,  so  the  coefficient  for  chords  AB  and  BC  is  4. 

Taking  moments  about  C  for  stress  in  top  chord  be,  we  have 
(4:W)2d/h-Wd/h,  so  the  coefficient  for  chord  be  is  7.  It  is  the  same  for 
the  bottom  chord  CD,  as  the  center  of  moments  in  that  case  is  at  c,  which 
is  in  the  same  vertical  line  as  point  C.  Taking  moments  about  either  d 
or  D,  for  the  stress  in  chords  cd  and  DE,  we  have  the  4  at  A  multiplied 
by  3d  minus  the  W  at  B  multiplied  by  2d  and  minus  the  W  at  C  multi- 


DESIGN  OF  SIMPLE  EAILKOAD  BKIDGES 


337 


plied  by  Id,  so  the  coefficients  for  chords  cd  and  DE  are  each  (4x3)- 
2-1  =  9. 

Similarly,  taking  moments  about  e  or  E,  for  the  stress  in  chords  de 
and  EF,  also  ef,  we  have  (4x4)  -  3  -  2  - 1  =  10  for  the  coefficient  to  be 
used  in  determining  the  stress  in  these  chords. 

After  a  little  practice  the  student  can  write  these  coefficients  down 
without  hesitating. 

If  the  slide  rule  is  used,,  about  all  that  need  be  written  down,  out- 
side of  the  stresses  (as  they  are  determined),,  are  the  numerical  values 
of  the  chord  coefficients  and  that  of  W,  tan  6,  and  sec  0,  and  if  found 
convenient  or  necessary,  W  tan  0  and  W  sec  0. 

174. — Determination  of  Live-Load  Stresses  in  Trusses. — First, 
draw  a  sketch  of  the  truss,  as  shown  in  Fig.  251,  for  reference. 

Member  bA.  Suppose  we  start  by  determining  the  maximum  stress 
in  the  end  post  bA.  As  is  readily  seen,  the  stress  in  this  member  will  be 


b  +  26OOOOC  +294000 d 


C  D  E  F 

e  Panels  @,  25-Q '=  /so -o" 


Fig.   251 

a  maximum  when  the  shear  in  the  panel  AB  is  a  maximum,  and  this  will 
occur  when  the  wheels  are  in  the  position  that  most  nearly  satisfies  the 
criterion  of  Art.  90,  which  is:  The  load  in  the  panel  AB  must  be  as 
nearly  equal  as  possible  to  the  total  load  on  the  bridge  divided  by  the 
number  of  panels,  and,  at  the  same  time,  a  wheel  must  be  at  B.  Now 
placing  wheel  4  at  B,  as  shown  in  Fig.  252,  we  have  (using  Table  A) 
125  -  91  =  34  ft.  of  uniform  load  on  the  bridge,  making  a  total  of  352,000* 
=  (284  +  34x2)1,000  per  truss,  and  dividing  by  the  number  of  panels 
we  have 


The  load  in  the  panel  including  one-half  of  wheel  4  is  60,000*.  So  this 
position  comes  as  near  to  satisfying  the  criterion  for  shear  as  is  possible, 
as  will  be  found  by  trying  the  other  positions. 


D 
Fig.   252 


34-' 


Taking  moments  about  G  (Fig.  252),  we  have  (using  Table  A) 


R=(  16,364  +  284x34  +  2 


!> 

2  /   ] 


OOP 
150 


=  181,172  Ibs. 


for  the  total  reaction  at  A.     Then  taking  moments   about  B,  we  hare 

,000  =  19,200  Ibs. 


r= 


338  STRUCTURAL  ENGINEERING 

for  the  floor  beam  concentration  at  A.     So  we  then  have 

tf-r=  181,1 72 -19,200  =  161,972,  say  162,000  Ibs. 

for  the  maximum  shear  in  the  end  panel  AB,  due  to  Cooper's  £40  loading 
and  for  E 50  we  have 

162,000x^5  =  902,500  Ibs. 

Then  for  the  maximum  live-load  stress  in  the  end  post  bA  we  have 
202,500x1.3  =  263,250,  say  263,000  Ibs.  (compression) 

Member  bB.  The  maximum  live-load  stress  on  the  hanger  bB,  as  is 
obvious,  is  equal  to  the  maximum  floor  beam  concentration  found  in  Art. 
171  to  be  94,500*,  and  hence  we  will  not  recalculate  it. 


Tte 


Fig.   253 


Member  bC.  The  maximum  stress  in  the  diagonal  bC  will  occur 
when  the  shear  in  panel  BC  is  a  maximum,  as  this  member  carries  the 
shear  in  that  panel.  Placing  wheel  3  at  panel  point  C,  as  shown  in  Fig. 
253,  we  have  100-96  =  4  ft.  of  uniform  load  on  the  bridge,  making  in  all 
292,000*=  (284  +  2x4)  1,000  on  one  truss. 

Dividing  this  by  the  number  of  panels,  we  have  292,000  -f-  6  =  48,666*, 
and  the  load  in  panel  BC  is  40,000*,  including  one-half  of  the  wheel  3, 
at  C  (see  Fig.  253).  This  does  not  seem  to  satisfy  the  criterion  very 
closely,  so  we  will  try  wheel  4  at  C,  in  which  case  we  have  9  ft.  of  uni- 
form load  on  the  bridge,  making  a  total  load  of  302,000*  per  truss,  and 
dividing  by  the  number  of  panels  we  have  302,000  -f  6  =  50,333*,  and  the 
load  in  the  panel  is  60,000.  This  position  does  not  satisfy  the  criterion 
as  closely  as  the  first,  and  hence  the  first  position,  with  wheel  3  at  C, 
will  be  taken.  It  is  evident  that  no  other  positions  need  be  tried,  as  the 
load  in  the  panel  in  the  first  case  was  too  light  and  in  the  second  it  was 
too  heavy. 

Then  taking  moments  about  G  (Fig.  253),  we  have  (using  Table  A) 


R  =  f  1  6,364  +  284  x  4  +  4^  ^jj?  =  116,773 
for  the  reaction  at  A,  and  taking  moments  about  C  we  have 

=9,200  Ibs. 


for  the  concentration  at  point  B. 

Then,  for  the  maximum  shear  in  the  panel  BC,  due  to  the   £40 
loading,  we  have 

R  -  r  =  116,773  -  9,200  =  107,573, 

and  for  the  £50  loading  we  have  134,466,  say  134,500*.     Then  multi- 
plying this  by  sec  B,  we  have 

134,500x1.3  =  174,850,  say  175,000  Ibs. 
for  the  maximum  tensile  stress  in  the  diagonal  bC. 


DESIGN  OF  SIMPLE  EAILKOAD  BKIDGES  339 

Member  cD.  The  maximum  live-load  stress  in  the  diagonal  cD  will 
occur  when  the  live-load  shear  in  the  panel  CD  is  a  maximum.  To 
satisfy  the  criterion  for  maximum  shear  in  that  panel,  let  us  try  wheel  3 
at  point  D  (see  Fig.  254).  This  position  of  the  wheels  brings  wheel  15 
exactly  at  G,  the  right  support. 


Fig.   254 

Then  the  load  on  the  bridge  (not  including  wheel  15)  is  232,000*. 
Dividing  this  by  the  number  of  panels,  we  have  232,000  -=-6  =  38,666#, 
and  the  load  in  the  panel  is  40,000*  (including  one-half  of  wheel  3). 
This  comes  as  near  satisfying  the  criterion  as  is  possible,  as  will  be  seen 
by  trying  other  positions. 

Now,  by  taking  moments  about  G  (using  Table  A)  we  obtain 


R  =  10,816  x  =  72,107  Ibs. 

lot) 

for  the  reaction  at  A,  and  taking  moments  about  D  we  obtain 

r  =  230  xi^  =  9,200  Ibs. 
^5 

for  the  concentration  at  C. 

Then,  for  the  shear  in  panel  CD,  due  to  the  £40  loading,  we  have 

R  -  r  =  72,107  -  9,200  -  62,907  Ibs. 

and  for  the  E50  loading  we  have  78,633#.     Multiplying  this  by  sec  0, 
we  have 

78,633  x  1.3  =  102,223,  say  102,000  Ibs. 

for  the  maximum  live-load  tensile  stress  in  the  diagonal  cD. 

Member  cC.  As  the  live  load  is  applied  wholly  on  the  bottom  chord 
joints,  the  post  cC  will  have  a  maximum  compression  stress  when  the 
diagonal  cD  has  maximum  tension,  as  the  post  takes  only  the  vertical 
component  of  the  stress  in  that  diagonal.  The  maximum  vertical  com- 
ponent of  the  stress  in  the  diagonal  cD,  as  seen  above,  is  equal  to  the 
maximum  shear  in  the  panel  CD  and  given  there  as  78,633.  So  the 
maximum  live-load  compressive  stress  in  the  post  cC  is,  say,  78,000*, 
using  round  numbers. 

Member  dD.  There  is  no  live-load  stress  in  the  post  dD  at  all,  as 
there  is  no  member  (as  a  diagonal)  connecting  to  the  top  of  it  that  will 
resist  vertical  action.  For  the  sake  of  illustration,  let  us  assume  a  live- 
load  stress,  S,  in  this  post.  Then  the  components  of  this  stress  must  be 
taken  by  the  top  chords,  and  we  would  have 

S  cos  90°  =  C  =  0,  as  cos  of  90°  =  0. 

Members  eE  and  eD.  If  the  span  is  loaded  from  the  right  end  so 
that  maximum  shear  occurs  in  panel  DE,  the  post  eE  will  be  subjected  to 
maximum  live-load  tensile  stress  and  the  diagonal  eD  will  be  subjected 
to  maximum  compressive  stress. 


340  STRUCTURAL  ENGINEERING 

If  wheel  2  be  placed  at  E,  wheel  10  will  be  just  2  ft.  from  the  right 
end  of  the  span,  as  shown  in  Fig.  255,  and  the  total  load  on  the  bridge 
will  be  152,000*. 

Dividing  this  by  the  number  of  panels  we  have  152,0004-6  =  25,333*, 
and  the  load  in  the  panel  DE  is  equal  to  20,000*,  including  one-half  of 
wheel  2.  Now,  if  wheel  3  be  placed  at  E,  the  total  load  on  the  bridge 


Fig.  256 


will  be  152,000*.      Dividing  this  by  the  number   of  panels,   we   have 
152,0004-6-25,333*,  and  the  load  hi  the  panel  DE  is  then  40,000*. 
So  it  is  seen  that  the  position  of  the  wheels,  shown  in  Fig.  255,  comes 
nearest  to  satisfying  the  criterion  for  shear  in  panel  DE. 
Taking  moments  about  G  (Fig.  255),  we  obtain 


=  ^4, 


,632  +  152  x  2  .  =  32,906  Ibs. 

for  the  reaction  at  A,  and  taking  moments  about  E  we  obtain 


25 

for  the  concentration  at  D. 

Then,  for  the  shear  in  panel  DEf  due  to  the  E40  loading,  we  have 

R  -  r  =  32,906  -  3,200  =  29,706  Ibs., 

and  multiplying  by  50/40  we  have  37,100*  for  the  shear  due  to  the  £50 
loading,  which  is  the  maximum  live-load  tensile  stress  in  post  eE,  and 
multiplying  this  by  sec  0  we  obtain  48,230*,  say  48,000*,  for  the  maxi- 
mum live-load  compression  stress  in  diagonal  eD. 

r 


Fig.  256 

Member  fE.  Now  placing  wheel  2  at  F,  wheel  6  will  be  just  1  ft. 
to  the  left  of  the  right  end  of  the  span  as  shown  in  Fig.  256.  Then  the 
load  on  the  span  is  103,000*.  Dividing  this  by  6  we  have  17,166*,  and 
the  load  in  panel  EF  is  20,000*.  This  position  comes  the  nearest  to 
satisfying  the  criterion,  as  will  be  found  by  trial. 

Taking  moments  about  G  (Fig.  256),  we  obtain 


R  =(l,640  + 103^^?  =  H,620  Ibs. 

\  / 

for  the  reaction  at  A,  and  taking  moments  about  F  we  obtain 

r  =  80x^5^  =  3,200  Ibs. 
for  the  concentration  at  E. 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  341 

Then,,  for  the  shear  in  panel  EF,  due  to  the  E5Q  loading,  we  have 


R  -  r~  =    ll,62Q  -  &20<j~  =  10,525  Ibs. 

Multiplying  this  by  sec  0  we  have  13,682*,  say  14,000*,  for  the 
maximum  live-load  compressive  stress  in  diagonal  fE. 

It  is  seen,  from  Fig.  251,  that  diagonal  bC  has  175,000*  tension  in 
it,  while  the  corresponding  diagonal  fEt  on  the  right  half  of  the  span, 
has  14,000*  compression,  that  the  post  cC  has  78,000*  compression, 
while  the  corresponding  post  eE  has  37,000*  tension  and  diagonal  cD  has 
102,000*  tension  while  the  corresponding  diagonal  on  the  right  half  of 
the  span  has  48,000*  compression.  Now  it  is  evident  that,  if  the  span 
is  loaded  from  the  left  end  instead  of  the  right  end,  as  it  was  above,  and 
the  maximum  stresses  in  the  web  members  computed,  just  the  opposite 
results  will  be  obtained,  that  is,  for  example,  diagonal  bC  would  be  sub- 
jected to  14,000*  compression,  while  diagonal  fE  would  be  subjected 
to  175,000*  tension. 

So  it  is  seen  that  we  have  now  determined  both  maximum  tensile 
and  compressive  live-load  stresses  in  the  web  members  throughout  the 
structure,  and  we  will  next  determine  the  chord  stresses. 

Members  AB  and  BC.  It  is  obvious  that,  if  either  of  the  members 
AB  or  BC  (Fig.  251)  be  cut,  rotation  would  take  place  instantly  about 


C  D  £  '  /=• 

Fig.  257 

joint  b.  Therefore  these  members  undoubtedly  prevent  rotation  about 
that  joint,  and  hence  the  greater  the  tendency  of  rotation  the  greater  the 
stress  in  these  two  members ;  so  undoubtedly  the  maximum  live-load  stress 
in  them  will  occur  when  the  live-load  bending  moment  about  b  is  a 
maximum. 

According  to  Art.  91,  the  moment  about  b  will  be  a  maximum  when 
the  average  unit  load  on  the  left  of  the  panel  point  is  equal  to  the  average 
unit  load  on  the  bridge. 

Placing  wheel  4  at  B  (Fig.  257),  we  have  34  ft.  of  uniform  load  on 
the  bridge  (see  Table  A).  Then  for  the  average  unit  load  on  the  span 
we  have 

(  284  +  34  x2\^^-  =  2,346  Ibs. 
\  j    lou 

and  for  .the  average  unit  load  on  the  left  of  B  (or  b)  we  have5  including 
one-half  of  wheel  4,  60,000  -r  25  =  2,400*. 

This  position  of  the  wheels  satisfies  the  criterion  as  nearly  as  pos- 
sible, as  will  be  found  by  trying  other  positions  of  the  wheels. 

Now,  taking  moments  about  G,  we  have 

R  =  181,173  Ibs. 


342 


STRUCTURAL  ENGINEERING 


for  the  reaction  at  A.     Then  taking  moments  about  b  we  have,  using 
Table  A, 

181,173  x  25  -  480  x  1,000  =  4,049,300  ft.  Ibs. 

for  the  maximum  bending  moment  about  b,  and  multiplying  this  by  50/40 
and  dividing  by  the  height  of  span  (see  Art.  92),  we  obtain  168,000* 
tension  for  the  maximum  live-load  stress  in  member  AB,  and  also  in  BC. 
Members  be  and  CD.  By  imagining  the  top  chord  be  (Fig.  258)  to 
be  cut,  it  is  readily  observed  that  this  member  resists  rotation  about  joint 
C.  So  the  live-load  stress  in  the  top  chord  be  will,  evidently,  be  a  maxi- 
mum when  the  load  is  in  the  position  for  maximum  moment  about  panel 
point  C,  and  a  maximum  in  the  bottom  chord  CD  when  the  load  is  in  the 
position  for  maximum  moment  about  panel  point  c.  But  as  joints  C  and  c 
are  in  the  same  vertical  line  the  position  of  the  wheels  will  be  the  same 
in  the  two  cases,  and  it  will  make  no  difference  which  joint  be  taken  as 
the  center  of  moments. 


28' 


B 
50' 


D 
Fig.  258 


By  placing  wheel  7  at  panel  point  C,  there  will  be  28  ft.  of  uniform 
load  on  the  bridge,  as  shown  in  Fig.  258,  and  we  have 

(284,000  +  56,000)  -f  150  =  2,266  Ibs. 
for  the  average  unit  load  on  the  bridge,  and 


103,000 


50  =  2,190  Ibs. 


for  the  average  unit  load  to  the  left  of  joint  C.  This  position  of  the 
wheels  comes  as  near  to  satisfying  the  criterion  for  maximum  moment 
about  joint  C  or  c  as  possible.  Then,  taking  moments  about  G,  we  have 
(using  Table  A) 

R=  ( 16,364 +  284x28 +  281 


^f  =  167,333  Ibs. 
loU 


for  the  reaction  at  A  (Fig.  258),  and  then  taking  moments  about  C 
we  have 

167,333  x  50  -  (2,155  x  1,000)  =  6,211,650  ft.  Ibs. 

for  the  bending  moment  about  joint  C.  Now,  it  is  a  question  as  to  Whether 
the  actual  maximum  moments  occur  at  C  or  at  the  corresponding  joint  E 
on  the  right  half  of  the  structure,  as  the  load  passes  to  the  left  over  the 
bridge.  Therefore,  to  make  sure,  we  will  test  for  joint  E. 

By  placing  wheel  14  at  E,  we  have  20  ft.  of  uniform  load  on  the 
bridge,  as  shown  in  Fig.  259,  and  for  the  average  unit  load  on  the  span 


DESIGN  OF  SIMPLE  RAILKOAD  BKIDGES 


343 


we  have 


^284  +  40  )i^  =  o,iGO, 


and  for  the  average  unit  load  to  the  left  of  E  we  have 


\ 


D 
Fig.   259 


This  position  of  the  wheels  comes  nearest  to  satisfying  the  criterion  for 
maximum  moment  about  E.  Taking  moments  about  G,  with  the  wheels 
in  this  position,  we  obtain 


=     16,364  +  284  x  20  + 


=  149,600 


for  the  reaction  at  A,  and  taking  moments  about  E  of  the  forces  to  the 
left  we  obtain 

149,600  x  100  -  8,728  x  1,000  =  6,232,000  ft.  Ibs. 

for  the  maximum  moment  about  E,  which  is  greater  than  found  above  for 
point  C,  and  consequently  will  be  taken  as  the  maximum  bending  moment 
for  determining  the  stress  in  chords  be  and  CD  or  for  the  corresponding 
chords  ef  and  DE  on  the  right  half  of  the  bridge — which  amounts  to  the 
same.  Multiplying  this  moment  by  50/40  and  dividing  by  30,  the  depth 
of  the  truss  (in  feet),  we  have  259,666,  say  260,000#,  for  the  stress  in 
top  chord  be  and  also  in  bottom  chord  CD,  being  compression  in  be  and 
tension  in  CD. 

It  is  obvious  from  Fig.  259  that  the  bending  moment  about  E  would 
be  increased  if  a  uniform  load  be  placed  to  the  left  of  wheel  1,  in  which 
case  a  uniform  load  would  be  preceding  the  engines  as  well  as  following 
them.  This  loading  is  used  in  some  cases,  but,  when  such  is  used,  the 
impact  added  should  be  materially  less  than  when  the  engines  are  not 
preceded  by  a  uniform  load,  for  the  reason  that  the  speed  of  the  train 
is  not  likely  to  be  very  high  when  the  engines  are  pushing  as  well  as 
pulling  a  load.  In  other  words,  a  train  so  loaded  is  not  likely  to  reach 
the  "critical  speed." 

As  an  illustration,  let  us  assume  wheel  13  at  E  and  18  ft.  of  uniform 
load  (2,000*  per  ft.)  ahead  of  the  engines,  as  indicated  in  Fig.  260. 

Then,  for  the  average  unit  load  on  the  bridge,  we  have 


36  +  284  +  30 


1,000 


=  2,333  Ibs., 


344 


STRUCTURAL  ENGINEERING 


and  for  the  average  unit  load  to  the  left  of  E  we  have 


36  +  202 


\  1,000 
/  100 


=  2,380  Ibs. 


This   position   of  the   loading  comes   nearest  to   satisfying  the   criterion 
for  the  maximum  moment  about  E.    Taking  moments  about  G,  we  obtain 


R  =  ( 36  x  141  + 16,364  +  284  x  15  + 
for  the  reaction  at  A. 


=  172,833  Ibs. 


Now  taking  moments  about  E,  of  the  forces  to  the  left,  we  obtain 
172,833  x  100  -  7,668  x  1,000  -  36  x  91  x  1,000  =  6,339,300  ft.  Ibs., 

for  the  maximum  bending  moment  due  to  Cooper's  £40  loading.  Multi- 
plying by  50/40  and  dividing  by  30  we  obtain  264,137#,  say  264,000*, 
for  the  stress  in  chords  be  and  CD,  which,  as  is  seen  from  the  above,  is 
about  4,000*  greater  than  for  the  usual  load,  but  if  the  impact  is  reduced, 
say,  50  per  cent  below  the  usual  amount,  the  final  result  would  be  much 
less  than  that  obtained  from  using  the  usual  loading. 

Member  cd.  It  is  evident  that  the  maximum  live-load  stress  will 
occur  in  the  top  chord  cd  when  the  live-load  moment  about  joint  D  is  a 
maximum.  By  placing  wheel  12  at  D  we  have  35  ft.  of  uniform  load,  as 
shown  in  Fig.  261,  and  for  the  average  unit  load  on  the  bridge  we  have 

=  2,360  Ibs, 

and  for  the  average  unit  load  on  the  left  of  D  we  have  (including  one- 
half  of  wheel  12) 


This  position  of  the  loading  comes  the  nearest  to  satisfying  the  crite- 
rion for  maximum  moment  about  joint  D.  Taking  moments  about  G,  we 
obtain 


R= 


5,364  +  284  x  35  +  35^^^  =  183,526  Ibs. 
lou 


DESIGN  OF  SIMPLE  KAILKOAD  BRIDGES 


345 


for  the  reaction  at  A,  and  taking  moments  about  joint  D  we  obtain 

183,526  x  75  -  6,708  x  1,000  =  7,056,450  ft.  Ibs. 

for  the  maximum  live-load  moment  about  that  point,  due  to  the  £40  load- 
ing. Then  multiplying  this  by  50/40  and  dividing  by  30  we  have  294,018, 
say  294,000*  compression  stress  in  top  chord  cd. 

As  the  bridge  is  symmetrical  about  the  center  of  span  we  have  all 
of  the  live-load  stresses  now  determined  and  we  will  next  determine  the 
impact  and  make  a  summary  of  all  the  stresses  in  the  trusses. 

175.  Summarizing  of  Stresses  and  Determination  of  Impact  in 
the  Trusses. —  The  dead-  and  live-load  stresses  can  be  written  on  the 
members,  as  shown  in  Fig.  262,  directly  from  Figs.  249  and  251,  and  we 
have  only  the  impact  to  determine. 

+  88000*0 
+  2GOOOO*L 
+  /7300O  I 
+3Z/OOO*  C 


—    55OOOD  B       -    55OOO*D 

-I  68OOO*L  -/68OOO*L 

-I  I  2000*1  -/  I  2000*1 

335000*  335000*  /  -52/OOO* 
6  Pane/s(§)  ZS-O=/BO'-  o" 


-  88OOO*D 

-  26OOOO*L 
-I  73000*1 


Fig.  262 


The  live  load  extends  practically  over  the  entire  length  of  the  bridge, 
as  seen  above,  when  the  chords  and  end  posts  receive  their  maximum  live- 
load  stress.  So,  in  determining  the  impact  stress  in  these  members,  L 
in  the  impact  formula  (Art.  125)  will  be  taken  as  the  total  length  of  the 
span,  or  150  ft. 


Then  we  have 


C  = 


300 


150  +  300 


=  0.6666 


for  the  coefficient  of  impact  for  these  members,  and  multiplying  the  live- 
load  stress  in  each  by  this  coefficient  we  obtain  the  impact  given  on  them 
in  Fig.  262.  As  an  example,  the  impact  in  the  end  post  bA  is  263,000  x 
0.6666  =  175,315,  say  175,000*;  in  chord  be  it  is  260,000x0.6666  =  173,- 
316,  say  173,000*;  and  so  on  for  the  other  chord  members. 

The  maximum  live-load  stress  in  hanger  bB  results  from  loads  in 
the  two  adjacent  panels,  AB  and  BC,  alone.  So  the  L  in  the  impact 
formula  would  be  taken  as  50  ft.,  the  sum  of  the  lengths  of  the  two  panels, 
in  determining  the  impact  stress  in  that  member.  Then  for  the  impact 
stress  in  hanger  bB  we  have 


When  the  maximum  live-load  tensile  stress  occurs  in  the  diagonal 
bC  the  live  load  extends  from  the  right  support  to  a  little  beyond  panel 


346  STRUCTURAL  ENGINEERING 

point  C  (see  Fig.  253).  In  the  case  of  a  uniform  live  load  the  distance 
from  the  right  support  C  would  be  taken  as  L  in  the  impact  formula,  as 
in  that  case  the  loads  are  considered  to  be  only  at  panel  points,  and  as  the 
result  is  affected  but  little  we  will  use  the  same  distance  in  the  case  of 
the  concentrated  loads.  So  for  the  impact  stress  in  diagonal  bC  we  have 

(300      \ 
100  +  300)  =131,250,  say  131,000  Ibs.  tension. 

The  maximum  live-load  compression  occurs  in  post  cC  when  the  load 
extends  from  the  right  support  to  a  little  beyond  panel  point  D,  and  the 
maximum  tension  occurs  in  diagonal  cD  (see  Fig.  254)  at  the  same  time. 
So  the  distance  from  the  right  support  to  panel  point  D  (75  ft.)  will  be 
taken  as  L  in  determining  the  impact  in  these  members.  Then  for  the 
impact  (compression)  in  post  cC  we  have 

78,000  x  (753+°3OQ)=6^400,  say  62,000  Ibs., 
and  for  the  impact  (tension)  in  diagonal  cD  we  have 

102,000  x  (75303°00)  =81,600,  say  82,000  Ibs. 

The  maximum  live-load  tension  in  post  cC  and  maximum  compression 
in  diagonal  cD  occur  when  the  load  extends  from  the  left  support  to  a 
little  beyond  panel  point  C.  (See  case  of  corresponding  members  (eE 
and  eD)  on  right  half  of  truss,  Fig.  255.)  So  the  distance  from  the  left 
support  to  panel  point  C  will  be  taken  as  L  in  determining  the  impact 
in  this  case,  and  we  have 

37,000  x  (35^55)  =31,685,  say  32,000  Ibs. 
for  the  impact  (tension)  in  post  cC  and 

48,000  x  (503+°3OQ)  =41,146,  say  41,000  Ibs. 

for  the  impact  (compression)  in  diagonal  cD. 

The  impact  stress  found  above  for  each  respective  member  can  now 
be  written  on  each  member,  and  adding  them  in  each  case  to  the  dead- 
and  live-load  stresses  the  total  combined  stresses,  as  shown  in  Fig.  262, 
are  obtained. 

In  the  case  of  members  cC  and  cD  we  have  what  is  known  as 
reversal  of  stress.  The  dead-load  stress,  which  acts  at  all  times,  is  com- 
pression in  post  cC  and  tension  in  diagonal  cD.  When  the  live  load 
moves  onto  the  structure  from  the  left  a  tensile  stress  of  37,000*  occurs 
in  post  cC  and  a  compression  stress  of  48,000*  in  diagonal  cD,  as  shown 
for  the  corresponding  members  (eE  and  cD)  on  the  right  half  of  the 
bridge  (Fig.  251).  Now  it  is  evident  that,  if  the  dead-load  compression  in 
post  cC  were  just  equal  to  the  37,000*  live-load  tension,  the  two  stresses 
would  just  balance  each  other  and  consequently  the  actual  stress  in  the 
post  would  be  zero  when  the  maximum  live-load  tension  occurred  in  the 
post;  but,  if  the  dead-load  stress  were  less  than  the  live,  the  actual  stress 
would  be  the  difference  between  the  two.  In  any  case,  before  the  dead 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  347 

load  is  subtracted  the  impact  should  be  added  to  the  live-load  stress.  To 
be  sure  that  reversal  is  well  provided  for,  as  a  rule  not  all  of  the  dead- 
load  stress  is  subtracted  from  the  sum  of  the  live-load  and  impact  stresses. 
The  A.  R.  E.  Ass'n  specification,  the  one  we  are  using,  calls  for  f. 
Therefore,  for  the  tensile  stress  in  post  cC,  we  have 

-3  7,000  -32,000  +  |x22,000  =  -54,334,  say  54,000  Ibs., 

o 
and  for  the  compressive  stress  in  diagonal  cD  we  have 

+48,000  +  41,000-|xl7,000  =  +77,667,  say  78,000  Ibs. 

o 

In  case  f  of  the  dead-load  stress  in  any  member  is  greater  than 
the  sum  of  the  reverse  live-load  stress  and  impact,  the  member  is  not 
reversed  and,  consequently,  the  reverse  stress  is  ignored,  as  in  the  case 
of  diagonal  bC.  In  bC  we  have  14,000*  (see  Fig.  251)  live-load  com- 
pression and  14,000  x  300  +(25  +  300)  =  13,000*  (about)  impact,  making 
in  all  a  stress  of  14,000  +  13,000  =  27,000*  compression,  but  f  of  the 
dead-load  tension  is  51,000x2/3  =  33,666*  and  hence  no  reversal  takes 
place,  that  is,  the  diagonal  is  always  in  tension,  due  to  dead  load,  and, 
consequently,  the  reversal  stress  can  be  ignored.  The  same  is  true  in  all 
such  cases. 

176.  Designing  of  Members  in  Trusses.—  After  all  of  the  maxi- 
mum stresses  are  computed  in  the  members  of  the  trusses  and  combined 
as  shown  in  Fig.  262,  the  next  thing,  logically,  to  do  is  to  design  the 
sections  of  these  members.  The  first  thing  to  do  in  such  work  is  to 
glance,  so  to  speak,  over  the  structure  and  associate  the  details  of  the 
different  members  and,  as  far  as  possible,  ascertain  the  governing 
members. 

In  this  case,  as  in  all  such  bridges,  the  intermediate  posts  are  the 
governing  members,  as  their  width  governs  the  width  of  practically  all 
of  the  other  members  of  the  trusses.  So  we  will  first  design  the  inter- 
mediate post  cC  (Fig.  262). 

Member  cC.  (See  Fig.  262.)  Let  us  assume 
each  of  these  posts  to  be  composed  of  2  —  15"  [s 
placed,  in  reference  to  each  other,  as  shown  in  Fig. 
263.  The  distance  d  between  the  toes  of  the  flanges 
is  really  the  governing  distance.  This  distance 
should  not  be  less  than  5£"  in  any  case,  and  a  little 
more  is  preferable.  If  this  distance  is  less  than 
5y  it  is  impossible  to  get  the  jaw  of  an  ordinary 
riveter  inside  of  the  member  and  consequently  the 

rivets  would  have  to  be  hand  driven,  which  is  expensive  work.  From 
table  3  it  is  seen  that  the  average  width  of  flange  of  15"  [s  is  about  3J", 
making  7"  for  the  two  flanges,  and  if  we  make  d=6J",  we  have 


for  the  width  of  the  posts.  The  least  average  radius  of  gyration,  which 
is  in  reference  to  axis  x-x,  as  seen  from  Table  3,  is  about  5.5".  Then 
substituting  this  value  of  r  in  the  column  formula  of  Art.  73,  and  taking 


348  STRUCTURAL  ENGINEERING 

L  as  30  ft.,  or  360  ins.,  we  have 

16,000 -70^  =  11,420  Ibs. 
o.o 

for  the  allowable  unit  compressive  stress  on  the  post. 

For  the  stress  in  the  post  we  have  162,000*  compression  and  54,000* 
tension,  as  given  in  Fig.  262.  Now,  according  to  the  specifications,  each 
of  these  stresses  must  be  increased  by  0.5  of  the  lesser.  So  we  have 
162,000  +  0.5x54,000  =  189,000*  compression  and  54,000  +  0.5x54,000  = 
81,000*  tension  which  the  post  must  be  designed  to  carry. 

Taking,  first,  the  case  of  compression,  we  have 

189,000 
11420   =16-6sq-ms-  (about) 

for  the  area  required  for  compression  and 

81,000 


16,000 


=  5.06  sq.  ins. 


for  the  net  area  required  for  tension.  The  area,  16.6n",  required  for 
compression  governs.  The  nearest  to  this  (considering  15"  channels)  is 
2— [s  15"x33*,  which  have  an  area  of  19.8n//.  These  channels  have 
3.2n"  more  area  than  required,  so  let  us  try  2— [s  12"x30*.  Then 
we  have 

16,000  -  70  252  - 10,100  Ibs. 

4.,«o 

for  the  allowable  unit  stress  on  the  post,  and  dividing  this  into  the  stress 
we  have 

189,000  +  10,100  =  18.7  sq.  ins. 

for  the  required  area.  This  shows  that  the  12"x30*  channels  are  too 
light  and  that  the  12"x35*  channels  would  have  to  be  used,  which  are 
heavier  than  the  15"x33*  channels;  and,  besides,  better  details  are  ob- 
tained by  using  the  15"  channels,  so  we  will  use  the  2 — [s  15"x33*. 
The  radius  of  gyration  of  these  channels  is  5.62,  within  0.12  of  the 
assumed  radius,  and  hence  recalculations  are  unnecessary. 

Member  dD.  The  post  dD  carries  nothing  but  the  9,000*  of  dead- 
load  compression  and  theoretically  could  be  made  of  very  light  section, 
but  to  keep  the  floor  beams  the  same  throughout  the  span  and  other  details 
constant,  as  well  as  for  general  appearance,  we  will  make  this  post  of 
2 — [s  15"x33*  (the  same  as  post  cC),  that  is,  each  of  these  posts  will 
be  made  of  2— [s  15"x33*  (the  lightest  15"  channels)  regardless  of 
the  excess  of  metal. 

Member  bB.  The  hangers  bB,  or  hip  verticals  as  they  are  often 
called,  are  not  posts  but  wholly  tension  members,  and  it  is  not  necessary 
that  they  be  of  the  same  type  of  section  as  the  posts;  however,  they  are 
often  made  so,  but  more  often  they  are  made  of  four  angles  built  into  an 
I-section.  (See  Fig.  298.)  The  total  stress,  as  given  in  Fig.  262,  divided 
by  the  allowable  unit  intensity  gives  193,500  +  16,000  =  12.09°"  (net) 
for  the  required  area  of  cross-section.  We  will  use  4 — Ls  6"  x  4"  x  f "  = 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  349 

14.44n"-1.5n"  =  12.94n"  (net),  deducting  a  one-inch  hole  (for  J-in. 
rivet)  out  of  each  angle.  This  type  of  member  is  cheaper  than  the  two 
channel  sections,  as  used  in  the  posts,  as  there  are  fewer  details  and,  con- 
sequently, less  shop  work,  but  it  is  not  an  economic  column  section,  as  the 
radius  of  gyration  is  small  in  comparison  with  the  area  of  cross-section 
contained,  as  in  comparison  with  two  channels  of  an  equal  cross-section. 
Member  bC.  The  diagonal  is  subjected  to  tension  only,  and,  using 
the  maximum  stress  given  in  Fig.  262,  we  have 

357,000  -=-  16,000  =  22.31  sq.  ins.  (net) 

for  the  required  area  of  cross-section.  We  could  use  either  four  angles, 
as  in  the  case  of  hanger  bB,  or  two  channels,  as  the  member  is  subjected 
to  tension  only,  but  the  two-channel  section  is  better  in  this  case  as  the 
member  is  subjected  to  some  cross  bending  due  to  its  own  weight,  and 
as  the  channel  section  is  the  more  capable  of  resisting  this  it  will  be  used. 
Let  us  try  2  —  [s  15"x40#.  We  can  consider  either  two  rivet  holes  cut 
out  of  each  web  or  one  hole  cut  through  each  flange,  whichever  is  the 
greater.  The  web  of  the  15"x40#  [  is  J£"  thick  (see  Table  3)  and 
the  grip  thickness  (g)  on  the  flange  is  §J",  hence  the  holes  through 
the  flanges  govern.  Then  we  have  2—  [s  15"  x  40*  =  23.52  -4  x  21/32  = 
20.90n".  As  is  seen,  these  two  channels  are  too  light.  So  let  us  try 
2  —  [s  15"x45#.  Jn  this  case  the  thickness  of  the  web  and  the  grip 
through  the  flanges  are  the  same,  and  hence  either  can  be  taken  in  deter- 
mining the  net  section  of  member.  So  we  have  2  —  [s  15"x45#  = 
26.48  -4xf  =  23.98°"  (net),  which  is  1.67n"  more  than  required,  but 
this  is  the  best  we  can  do  if  we  use  channels,  so  we  will  use  this  section 
for  diagonal  bC. 

Member  cD.  This  diagonal  is  subjected  to  201,000*  tension  and 
78,000*  compression,  as  given  in  Figv  262.  Now  increasing  each  by  0.5 
of  the  lesser,  as  we  did  above  in  the  case  of  post  cC  (according  to  the 
specifications),  we  have 

201,000  +  0.5  x  78,000  =  240,000  Ibs.  tension 

and 

78,000  +  0.5  x  78,000  =  117,000  Ibs.  compression. 

Then  dividing  the  tensile  stress  by  16,000*  (the  allowable  unit  stress) 
we  have 

240,000  -5-  16,000  =  15.0  sq.  ins. 

for  the  net  area  of  cross-section  required  to  carry  the  tension.  Now 
assuming  the  member  to  be  made  of  two  15"  channels  having  a  radius  of 
gyration  of  5.6  (a  mere  guess)  and  taking  the  length  of  the  diagonal  as 
39  ft.,  or  468  ins.,  we  have 


16,000  -70  y    =  10,150  Ibs. 

for  the  allowable  compressive  unit  stress.     Dividing  this  into  the  com- 
pressive  stress,  we  have 

117,000  -r  10,150  =  11.5  sq.  ins. 

for  the  required  area  of  cross-section  to  carry  the  compression.     Deduct- 
ing for  four  rivet  holes,  out  of  the  flanges  of  the  two  lightest  15"  channels, 


350  STRUCTURAL  ENGINEERING 

we  have  2—  [s  15" x  33#=  19.8-4x21/32  =  17.2n"  (about)  for  the  net 
area  to  carry  tension,  which  is  2.2n"  more  than  required.  The  total, 
19.8n//  is  available  for  compression,  so  the  section  is  larger  than  is  neces- 
sary for  compression  as  well  as  for  tension,  but  this  section  will  be  used 
as  it  is  desirable  to  have  all  of  the  diagonals  of  the  same  width.  We  now 
have  all  of  the  web  members  designed  and  we  will  next  take  up  the 
designing  of  the  chord  members. 

Members  AB  and  BC.  The  bottom  chords  are  purely  tension  mem- 
bers; so,  dividing  the  maximum  stress  in  chord  AB  or  BC  by  16,000, 
we  have 

335,000  -r  16,000  =  20.93  sq.  ins. 

for  the  net  area  required.  By  glancing  over  the  table  of  channels  (see 
Table  3)  we  see  that  two  15"  channels  can  be  used  as  section  for  these 
members.  Deducting  four  rivet  holes  from  the  flanges  we  have  2 — [s 
15"x40#  =  23.52-  2.6  =  20.9n",  which  is  practically  the  section  required, 
and  hence  these  channels  will  be  used. 

Member  CD.  For  the  net  area  of  cross-section  required  for  chord 
CD,  we  have  521,000 -f  16,000  =  32.56  sq.  ins.  Now,  from  the  table  of 
channels  (see  Table  3  or  any  structural  handbook,  as  Cambria  or  Car- 
negie) it  is  seen  that  no  two  15"  channels  will  have  the  required  net  sec- 
tion. So,  if  channels  are  used,  they  will  have  to  be  reinforced  by  riveting 
plates  to  their  webs.  Let  us  try  the  following  sections: 

2—  [s!5"x45#     =26.48-4x5/8    =  23.98n"net 
2— pis.  12"  x  7/16  =  10.50  -  4  x  7/16  =    8.75n"  net 

32.73n"  net 

The  net  area  of  this  is  practically  equal  to  the  area  of  the  required 
section  and  hence  will  be  used. 

Top  chords.  The  top  chords  and  end  posts  of  such  bridges  are  usu- 
ally made  up  of  either  two  channels  and  a  cover  plate  as  shown  at  (a), 

Fig.  264,  or  of  two  built  chan- 
nels  and  a  cover  plate  as  shown 
at  (6).  The  rolled  channels  are 
used  whenever  that  section  is 
sufficient.  It  is  not  desirable  to 
use  the  heaviest  channels,  which 
are  of  -Jf"  metal,  as  the  shop 
work  is  rather  expensive  on  ac- 
count of  the  thick  metal,  and 
whenever  the  end  post  or  heav- 

.  264  iest   chord  members   exceed   the 

area  of  the  cover  plate  and,  say, 

two  15"x45#  channels,  a  section  similar  to  the  one  shown  at  (6),  Fig. 
264,  is  used  throughout  the  structure,  as  it  would  be  unsightly  construction 
to  build  part  of  the  chord  members  of  a  bridge  of  rolled  channels  and 
part  of  plates  and  angles.  So,  in  this  case,  we  will  first  ascertain  as  to 
whether  we  can  or  cannot  use  the  rolled  channel  section  as  shown  at  (a), 
Fig.  264,  for  the  top  chord.  The  member  cd,  as  is  seen  from  Fig.  262,  will 
be  the  heaviest  member,  that  is,  the  one  requiring  the  greatest  area  of 


DESIGN  OF  SIMPLE  KAILEOAD  BEIDGES  351 

cross-section.  In  preliminary  calculations  the  least  radius  of  gyration  of 
this  type  of  section  can  be  taken  as  0.40  of  its  depth.  Then,  assuming  15" 
channels,  we  have  15x0.40  =  6.0  for  the  approximate  radius  of  gyration 
about  axis  x-x,  which  is  usually  the  least  radius.  Now  substituting  this 
value  of  r  and  the  length  of  the  member,  which  is  25  ft.,  or  300  ins.,  into 
the  column  formula,  we  have 


16,000  -7o        =  12,500  Ibs. 

0.0 

for  the  allowable  unit  stress  on  the  member.  Then  dividing  the  maximum 
stress,  as  given  in  Fig.  262,  by  this,  we  have 

589,000 

-j^-  =47.1  sq.  ins. 

for  the  required  area  of  cross-section  of  the  member.  The  width  of  the 
intermediate  posts,  as  found  above,  is  13J",  and  allowing  for  two  f" 
gusset  plates  and  7"  (  =  3-J-x2)  for  the  flanges  of  the  channels,  we  have 

13J  +  lJ  +  7  =  21f,  say  22  ins. 

for  the  width  of  the  cover  plate.  The  cover  plate  should  be  as  thin  as  is 
consistent  with  good  practice.  The  specifications  limit  this  to  -^  of 
the  distance  (shown  as  d  at  (a),  Fig.  264)  between  the  lines  of  rivets 
connecting  the  plate  to  the  other  parts  of  the  member.  That  distance 
here,  using  2"  gauge  in  channel  flanges,  is  13^  +  lJ  +  4  =  18  j",  say  19". 
One-  fortieth  of  this  is  practically  J",  so  the  cover  plate  would  be  22"  x-J". 
Now,  using  the  heaviest  channels  that  it  is  practical  to  use,  we  have  the 
following  sections  : 

2—  [s!5"x45#       =  26.48n" 

1—  cov.  pi.  22"  x  £"  =  11.00n" 
37.48*" 

This  is  about  10n"  less  than  the  required  area  found  above,  and  hence 
the  rolled  channel  section  has  not  sufficient  area;  and,  therefore,  the  type 
of  section  shown  at  (6),  Fig.  264,  will  be  used. 

In  designing  the  top  chords  it  is  best  to  begin  with  the  lightest  sec- 
tion, which  is  the  member  having  the  least  stress,  and  make  it  of  minimum 
thickness  of  metal  and  then  increase  the  area  of  cross-section  of  the 
other  top  chord  members  by  increasing  the  thickness  of  webs,  leaving  the 
other  parts  about  constant.  For  by  so  doing  the  centers  of  gravity  of 
the  chords  will  practically  be  in  the  same  plane  throughout.  Therefore, 
we  will  take  up  the  designing  of  chord  be  first  —  it  being  the  lightest. 

Member  be.  The  maximum  stress  in  chord  be,  as  given  in  Fig.  262, 
is  521,000*.  Then  assuming  the  allowable  unit  stress  to  be  12,500"  (see 
Example  4,  Art.  74),  we  have 

521,000 

^  41.68  sq.ms. 


tor  the  approximate  required  area.  Next  draw  a  sketch  of  the  cross- 
section  of  the  chord  as  shown  in  Fig.  265.  The  cover  plate  and  top 
angles  should  be  made  as  light  as  the  specifications  will  permit  in  order 
that  the  gravity  axis  x-x  be  as  near  the  center  of  the  web  as  possible.  As 
seen  above,  the  cover  plate  must  be  about  \"  thick  and  the  smallest  angles 


352 


STRUCTURAL  ENGINEERING 


that  can  be  used  at  the  top,  considering  details  and  all,  are  3J"  x  3J"  x  f  ". 
The  bottom  angles  should  be  as  heavy  as  is  practical  to  use  for  the  same 
reason  that  the  top  ones  are  made  light.  So  for  the  bottom  angles  we 
will  use  6"x4"xf",  as  that  is  about  the  maximum  size  and  thickness 
used  for  such  work.  The  cover  plate,  top  angles,  and  the  bottom  angles 
are  usually  about  constant  throughout  the  chord. 

Allowing  for  2  —  f  "  gusset  plates  and  for  2  —  |"  webs  (a  mere  guess), 
we  have 


for  the  width  of  cover  plate.     To  insure  a  neat  finish,  the  cover  plate 

should  project  J"  or  \"  beyond  the  angles.     So  we  will  make  the  cover 
^  plate  23"  wide.     However,  the  details 

should  always  be  such  that  the  width  of 
cover  plate  would  be  in  even  inches. 

The  depth  of  webs  should  be  such 
that  the  radius  of  gyration  about  the 
horizontal  axis  (x-x,  Fig.  265)  is  about 
equal  to  the  radius  about  the  vertical 
axis  (#-#).  For  preliminary  calcula- 
tions the  radius  of  gyration  in  reference 
to  axis  x-x  of  such  chords  can  be  taken 
as  0.4  of  the  depth  of  the  web,  and  the 
radius  in  reference  to  axis  y-y  can  be 
taken  as  the  horizontal  distance  out 

from  the  axis  to  the  outer  face  of  the  web.     So,  in  this  case,  we  have 

(assuming  \"  webs) 


</ 

t 

—  i- 
L 

JL 

^    I 

** 

—  -fr^ 

v,' 

t_ 

"N 

k 

<* 

\ 

* 

i     Jy 

i.se  V  e 

78' 

Fig.  265 


for  the  approximate  radius  about  axis  y-y.  Then  for  equal  radii  about 
the  two  axes,  we  have  0.4A  =  7.87,  from  which  we  obtain  for  the  depth 
of  the  web 


7.87 
-T—  - 
0.4 


1Q,. 
=19.  6  ins. 


As  the  work  here  is  only  fairly  approximate  this  figure  shows  only  about 
what  the  depth  of  the  web  should  be,  and  hence  any  web  about  this  depth 
can  be  used.  A  19"  web  is  an  odd  width  and  a  20"  web  appears  a  little- 
deep  for  this  section,  so  we  will  use  an  18"  web.  Now  the  radius  of 
gyration  of  the  section  about  the  horizontal  gravity  axis,  which  is  usu- 
ally the  least  radius,  is  about  0.4  of  the  depth  of  the  web,  as  stated  above, 
so  we  have 

0.4x18  =  7.2  ins. 

for  the  approximate  value  of  the  least  radius  of  gyration  of  the  section. 
Then  substituting  this  value  of  r  in  the  column  formula,  we  have 


16,000  -  70         =  13,084,  say  13,000  Ibs. 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES  353 

for  the  approximate  allowable  unit  stress.  Then  dividing  this  into  the 
stress  we  have 

521,000  +  13,000  =  40.07  sq.  ins. 

for  the  approximate  required  area.  Taking  this  as  a  guide,  let  us  assume 
the  following  section  for  chord  be: 

1— cov.  pi.  23"  xj"  =  11.50n" 
2— web  pis.  18"  x  f"  =  13.50n" 
2— Ls  3£"x3£"x£"  =  4.96°" 
2— Ls  6"x4"xTV'  =  10.62n" 
40.58°" 

The  next  thing  in  order  is  to  determine  the  center  of  gravity,  moment 
of  inertia  and  radius  of  gyration  of  this  section  and  check  back  to  see  if 
the  section  is  actually  the  correct  one  to  use. 

Taking  moments  about  the  center  of  the  cover  plate  (Fig.  265), 
we  have 

-      13.50  x  9.37  -f- 4.96  x  1.26  + 10.62  x  16.49 

-4^58~  -=7.59  ins. 

for  the  distance  from  the  center  of  the  cover  plate  to  the  horizontal  grav- 
ity axis  x-x,  and  for  the  moment  of  inertia  about  this  axis  we  have  the 
following : 

cover  plate  11.50  x  7^592  +  0  =    662.49 

webs  13.50x178%  182.25x2  =   407.27 

top  angles  4.96  x  673?  +  2.87  x  2  =   204.48 

bot.  angles   10.62  x  8^872  + 19.26  x  2     =    874.07 

2,148.31 

Then  for  the  radius  of  gyration  about  this  same  axis  x-x  we  have 


r  = 

which  is  within  0.08  of  0.4  of  the  depth  of  the  web. 

As  the  section  is  symmetrical  about  the  vertical  plane  through  the 
center  of  the  cover  plate,  the  vertical  gravity  axis  y-y  will  pass  through 
the  center  of  the  cover  plate  and  for  the  moment  of  inertia  of  the  section 
about  this  axis  y-y  we  have  the  following: 

cover  plate  0  +  506.96  =    506.96 

webs  13.50  x7J56a+0  =   771.57 

top  angles  4.96x8.76* +2.87x8   =   386.36 

bot.  angles  10.62x8.78  +6.91x2=   832.50 

2,497.39 


354  STRUCTURAL  ENGINEERING 

Then  for  the  radius  of  gyration  about  this  axis  y-y  we  have 


40.58 


=  7.84  ins., 


which  is  a  little  larger  than  the  radius  about  the  x-x  axis. 

Now  substituting  the  value  of  r   (the  least  radius)   in  the  column 
formula,  we  have 


16,000-  70  =  13,116  Ibs. 

7.6o 

for  the  actual  allowable  unit  stress  on  the  chord.     Dividing  this  into  the 
stress,  we  have 

521,000      ,-„„ 

=  39.72  «,.„,,., 


which  is  0.86n"  less  than  the  section  assumed  above,  but  as  this  is  about 
as  close  as  we  can  obtain  we  will  use  the  assumed  section  for  chord  be. 
Member  cd.  There  will  be  such  a  small  difference  between  the  allow- 
able unit  stress  for  the  other  top  chord  members  and  that  found  above 
for  chord  be  (as  can  be  verified  by  actual  calculations)  that  the  allowable 
unit  stress  found  for  that  member  will  be  used  in  designing  the  others. 
So,  dividing  the  maximum  stress  in  chord  cd,  as  given  in  Fig.  262,  by 
that  unit  stress,  we  have 

589,000 

=  44.9  sq.  ins. 


for  the  required  area  of  cross-section  of  the  member  cd. 

Then  using  the  same  size  angles  and  cover  plate  as  for  chord  be 
and  increasing  the  thickness  of  the  web  to  -J"  ',  we  have 

1—  cover  pi.  23"  x  \"  =11.50°" 

2—  web  pis.  18"  x  £"  -  18.00n" 
2—  Ls  3i"x3i"xf"  =   4.96n" 
2—  Ls  6"x4"xTV  =10.62°" 

45.08n" 

This  section  is  quite  close  to  the  required  section,  being  only  0.1  8n" 
larger,  arid  hence  will  be  used  for  chord  cd. 

This  completes  the  design  of  the  main  truss  members,,  as  the  bridge 
is  symmetrical  about  the  center  of  the  span,  except  the  end  posts,  which 
will  be  designed  later,  after  the  bending  moment  on  them  due  to  the 
wind  pressure  is  determined. 

177.  Designing  of  the  Bottom  Lateral  System.—  The  bottom 
lateral  system  is  really  a  horizontal  truss  in  the  plane  of  the  bottom  chord. 
It  is  for  the  purpose  of  resisting  the  wind  pressure  on  the  lower  portion 
of  the  structure,  and,  in  the  case  of  through  bridges  (such  as  the  one  we 
are  now  designing),  the  wind  pressure  on  the  train  as  well  as  the  vibra- 
tion due  to  the  swaying  of  the  train.  A  double  system  of  diagonals,  or 
laterals,  is  practically  always  used,  which  is  as  indicated  in  Fig.  266, 
where  the  bottom  lateral  system  is  shown  in  plain  view,  one  system  of 
laterals  being  dotted  to  avoid  confusion. 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES 


355 


The  bottom  chords  of  the  main  trusses  act  as  the  chords  in  this  lat- 
eral system  and  the  floor  beams  as  posts.  The  stress  in  the  chords  due 
to  the  wind  and  vibration  is,  as  a  rule,  ignored  unless  it  reverses  the  dead- 
load  stress,  taking  the  wind  pressure  in  that  case  on  the  unloaded  struc- 
ture as  350#  per  foot  of  span  or  where  the  maximum  exceeds  25  per  cent 
of  the  maximum  combined  dead,  live,  and  impact  stresses  (see  specifica- 
tions), in  which  case  it  is  combined  with  the  dead,  live,  and  impact 
stresses  as  in  the  case  of  viaduct  columns  (Art.  162).  This  combination, 
however,  is  very  rarely  necessary  in  ordinary  bridges.  The  stress  in  the 
floor  beams  (acting  as  columns),  due  to  wind  and  vibration,  is  ignored 
entirely,  as  the  laterals  are  connected  to  the  bottom  flange  of  the  beams, 


e  Bo  f.  Chord. 


6  Panefs  <&,  2S- o"=  ts 
Fig.  266 


which  are  in  tension  due  to  dead  and  live  load,  and  hence  the  compres- 
sion in  these  same  flanges,  due  to  wind  and  vibration,  only  tends  to 
reverse  the  dead-  and  live-load  stress  in  them. 

There  are  two  ways  of  considering  the  bottom  lateral  system:  One 
is  to  consider  the  double  system,  in  which  case  the  two  diagonals  in  any 
panel  resist  equally  the  shear  in  the  panel,  that  is,  one  diagonal  is  con- 
sidered as  having  the  same  intensity  of  stress  as  the  other,  one  being  in 
compression  and  the  other  in  tension;  the  other  way  is  to  consider  only 
a  single  system,  one  system  being  in  action  when  the  pressure  comes 
from  one  direction  and  the  other  system  acting  when  the  pressure  comes 
from  the  opposite  direction.  We  will  consider  the  single  system,  in  which 
case  the  diagonals  carry  only  tension,  and  then  investigate  for  rigidity 
after  the  sections  are  determined  for  the  single  system. 

According  to  the  specifications,  the  pressure  or  horizontal  load  on 
the  bottom  lateral  system,  here  considered,  is  200  +  0.10  x  5,000  =  700# 
per  foot  of  span.  So,  for  a  panel  load  we  have 

W=  700  x  25  =  17,500  Ibs. 

This  is,  according  to  the  specifications,  to  be  considered  a  moving  load 
(live  load).  Let  us  assume  that  this  load  acts  in  the  direction  indicated 
by  the  arrows  (Fig.  266)  and  that  it  moves  over  the  structure  from  right 
to  left,  that  is,  from  G  to  A.  Now,  from  Art.  90,  we  have 


=^Fl  +  2....  (n-l)l 


for  the  maximum  shear  in  the  different  panels,  and  hence  for  the  maximum 
stress  in  the  corresponding  diagonal  (considering  single  system)  we  have 


sec  W. 


356  STRUCTURAL  ENGINEERING 

Tan  a)  =  25/16  =  1.56,  and  sec  <o  =  1.85,  which  is  most  readily  obtained 
from  a  table  of  natural  functions  after  the  tangent  is  computed. 
Then  for  the  maximum  stress  in  diagonal  cD  we  have 

S=—(I  +  2  +  3)  sec  eo  =  17^°°  x  6  x  1.85  =  32,375, 
n  o 

say  32,000  Ibs.  (tension), 
and  for  the  maximum  stress  in  bC  we  have 

S'=—  (1  +  2  +  3  +  4)  seco)=i^|—  x  10x1.85  =  53,958, 
n  o 

say  54,000  Ibs.  (tension), 
and  for  the  maximum  stress  in  aB  we  have 


S"  =  —  (1  +  2  +  3  +  4+5  )secco  =  —  '-        x  15  x  1.85  =  80,937, 
n  o 

say  81,000  Ibs.  (tension). 

This  completes  the  necessary  determination  of  stress  in  the  diagonals  or 
laterals  (as  they  are  called)  due  to  wind  and  vibration,  as  the  structure 
is  symmetrical  about  the  center  of  the  span,  and  we  will  next  design  the 
sections  for  these  members. 

Taking  the  first  lateral  aB,  we  have 

81,000 


for  the  net  area  of  cross-section  required.  Using  1  —  L  5"x3|"xf"  = 
5.81-0.75  =  5.06n//,  we  have  exactly  the  correct  net  area  for  that  diag- 
onal. For  lateral  bC,  we  have 

54,000 


for  the  net  area  of  cross-section  required  and  by  using  1  —  L  5"  x  3  J"  x  J" 
=  4-0.5  =  3.5n",  we  have  about  the  correct  area,  at  least  about  as  near 
the  correct  area  as  is  possible  to  obtain.  For  lateral  cD,  we  have 

32,000 


for  the  net  area  of  cross-section  required,  and  by  using  1  —  L  5"  x  3|"  x  f  " 
=  3.05-0.37  =  2.68n"  we  have  0.68n"  more  net  area  than  required,  but 
we  will  use  this  angle  so  as  to  have  the  bottom  laterals  made  of  5"  x  3J" 
angles  throughout. 

As  the  structure  is  symmetrical  about  the  center  of  the  span,  we 
now  have  all  of  the  laterals  designed,  considering  the  single  system,  and 
we  will  now  investigate  the  double  system.  The  laterals  are  connected 
to  the  bottom  of  the  stringers  at  the  points  of  intersection  and  hence 
the  longest  unsupported  length  of  lateral  is  between  the  stringer  and 
truss.  This  maximum  unsupported  length  is  about  8'-0",  or  96".  Now 
taking  the  end  lateral  aB,  which  is  composed  of  1  —  L  5"  x  3|"  x  J",  we 
have  L/r  =  96/0.96  =  100,  which  is  quite  satisfying  as  the  maximum  value 


DESIGN  OF  SIMPLE  KAILROAD  BEIDGES 


357 


for  L/r  is  120,  and  substituting  the  value  of  L  and  r  in  the  column 
formula  we  have 

16,000  -70-^-  =  9,000  Ibs. 

\)  •  \)  O 

for  the  allowable  compressive  unit  stress,  and  dividing  this  into  one-half 
of  the  stress  found  above  (the  stress  that  it  would  be  considered  to  resist 
if  acting  in  the  double  system)  we  have 

40,500 


for  the  required  area  of  cross-section,  which  is  considerably  less  than  the 
cross-section  of  the  angle.  So  this  lateral  is  sufficient,  considering  either 
a  single  or  double  system,  and  the  same  is  true  for  the  other  laterals,  as 
will  be  found  upon  investigation.  So  the  bottom  laterals  as  designed  are 
capable  of  resisting  the  forces  coming  on  them  from  wind  and  vibration 
when  considered  either  as  a  single  or  double  system. 

The  bottom  laterals  are  subjected  to  stress  from  traction  in  addition 
to  the  stress  resulting  from  wind  and  vibration.  As  an  illustration,  sup- 
pose a  rapidly  moving  train  to  come  onto  the  structure  and  the  brakes  to 


floor  Beam*.           Bottom  Chord-*              Stringer-* 

( 

I 

A  I 

~^« 

/ 

^  /  1 

/ 

1 

Fig.  267 

be  suddenly  applied  —  the  wheels  would  skid  or  tend  to  skid  along  the 
rails,  whereby  a  longitudinal  thrust  would  be  exerted  along  the  stringers 
which  would  tend  to  bend  the  floor  beams  transversely  as  indicated  in 
Fig.  267  (where  the  train  is  assumed  to  be  moving  from  right  to  left). 
Now  this  bending  of  the  floor  beams,  which  would  cause  serious  stresses 
in  them,  is  prevented  by  attaching  the  laterals  to  the  bottom  of  the  string- 
ers so  that  the  longitudinal  thrust  exerted  by  the  stringers  is  transmitted 
directly  to  the  laterals,  and  from  there  to  the  bottom  chords  of  the  trusses 
and  on  out  to  the  end  supports.  As  the  stringers  are  connected  to  one 
another  rigidly,  end  to  end,  throughout  the  length  of  the  structure,  it  is 
evident  that  the  traction  from  any  loading  would  be  transferred  more  or 
less  from  stringer  to  stringer,  and  hence  would  be  distributed  to  some 
extent  to  all  of  the  laterals;  so  in  computing  the  stress  in  the  laterals 
due  to  traction  we  will  consider  that  the  laterals  in  any  panel  resist 
the  traction  from  an  average  panel  load  of  fully  loaded  bridge.  Then 
placing  wheel  1  (see  Table  A)  at  one  end  of  the  bridge,  we  have 


for  the  weight  of  the  full  live  load  (Cooper's  E50  loading)  on  the  bridge. 
Dividing  this  by  the  number  of  panels,  we  have 


6 


=  153,500  Ibs. 


358 


STRUCTURAL  ENGINEERING 


Stringer 


I52BO* 


for  the  average  panel  load.  Two-tenths  (0.2)  of  this  vertical  load  is 
considered  as  traction  and  is  exerted  along  the  track,  that  is,  the  coeffi- 
cient of  friction  of  the  wheels  on  the  rails  is  taken  as  0.2.  So  we  have 

152,500  x  0.2  =  30,500  Ibs. 

as  the  traction  force  exerted 
on  the  two  stringers  in  each 
panel,  or  15,250*  to  each 
stringer.  This  force  on  each 
stringer  is  transmitted  to 
the  laterals  at  two  points, 
where  the  laterals  and 
stringers  connect,  so  we  can 
consider  the  case  as  indi- 
cated in  Fig.  268.  One 
component  of  the  traction 
force,  15,250#  -f  2  applied  at 
each  of  the  points  c  will,  in 
each  case,  act  along  the  lat- 
eral, causing  compression  in 
the  case  of  the  part  Be  and  tension  in  the  part  DC,  and  the  other  com- 
ponent at  each  point  c  will  act  along  member  cc. 

Without  the  members  cc,  there  would  be  transverse  bending  on  the 
stringers  from  traction,  and  hence  cc  is  an  essential  member  in  the 
system.* 

Then  for  the  stress  in  a  lateral,  due  to  traction,  we  have 


do. 


Fig.   268 


15,250 


xsec</>  = 


15,250 


x  1.18  =  8,997,  say  9,000  Ibs., 


which  is  compression  in  the  part  cB  and  tension  in  part  cD,  and  just  the 
reverse  if  the  train  were  moving  in  the  opposite  direction.  For  the  stress 
in  member  cc,  which  can  be  called  a  tie  or  strut,  we  have 


15,250 


x  tan  <j>  = 


15,250 


x  0.64  =  4,880  Ibs., 


which  is  compression  in  one  and  tension  in  the  other. 

These  stresses  are  not  very  great  in  the  first  place,  as  is  seen,  and 
they  are  not  very  likely  to  occur  at  the  same  time  that  the  maximum 
wind  stress  occurs  and  hence,  as  a  rule,  are  ignored  in  the  designing  of 
the  sections  of  the  laterals.  However,  the  traction  should  be  provided 
for  in  so  far  as  making  the  laterals  capable  of  taking  compression 
(L/r<120)  and  the  inserting  of  the  ties  cc.  The  tie  cc  does  not  neces- 
sarily have  to  be  a  very  heavy  section.  They  are  usually  made  of  1 — L 
3J"x3"x£",  in  which  case  we  have 

7fi 
16,000 -70^  =  9,934  Ibs. 


*  As  far  as  the  author  knows,  Dr.  J.  A.   L.  Waddell,  M.A.Soc.C.E.,    was   the  first  to 
point  out  the  necessity  of  these  struts. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  359 

for  the  allowable  unit  stress  for  compression.     Then  we  have 

4,880 


for  the  required  area  in  the  case  of  compression,  and  less  is  required  for 
tension,  as  is  obvious.  So  it  is  seen  that  the  theoretical  required  area  of 
these  struts  (cc)  is  not'  the  governing  feature,  but  that  it  is  a  matter  of 
obtaining  sufficient  rigidity.  That  means  that  L/r  should  not  exceed  120 
in  any  case.  The  L/r  for  a  3^"x3"xf"  angle  is  about  0.87,  which  is 
reasonably  low.  This  angle  is  used  as  it  is  a  very  common  angle  in  this 
class  of  work,  and  it  is  about  the  minimum  size  permitted  and  it  fits  the 
case  wherein  common  judgment  decides. 

For  the  wind  stress  in  the  bottom  chord  ab  or  BC  (Fig.  266)  we  have 

2J  W  tan  (o  =  2J  x  17,500  x  1.56  =  68,250  Ibs., 

which  is  compression  in  ab  and  tension  in  BC,  in  this  case.  Now,  as 
shown  in  Fig.  262,  the  maximum  combined  stress,  dead,  live  and  impact, 
is  335,000*  tension.  Twenty-five  per  cent  of  this  is  83,750*,  so  the  maxi- 
mum wind  stress  in  these  chords  can  be  ignored.  (See  specifications.) 
In  case  of  combined  dead,  live,  impact,  and  wind  stresses,  the  allowable 
unit  stress  is  raised  25  per  cent,  as  the  probability  of  all  of  these  stresses 
being  a  maximum  at  the  same  time  is  very  remote.  The  same  is  true 
of  the  other  bottom  chords,  as  will  be  found  upon  investigation.  Now, 
so  far,  the  wind  stresses  in  the  bottom  chords  can  be  ignored  ;  but  further, 
suppose  the  bridge  to  be  unloaded,  that  is,  no  live  load  on  it,  the  wind 
pressure  of  course  will  be  less.  Let  us  then  assume  a  pressure  of  350* 
per  foot  of  span,  which  is  about  correct,  instead  of  700*.  Then  the 
stress  in  the  chord  ab  or  BC  is 


68,250  x         =  34,125  Ibs., 


which  here,  in  the  case  of  ab,  is  compression.  But,  as  this  does  not 
reverse  the  dead-load  tension  in  the  member,  which  is  55,000*,  the  wind- 
load  stress  need  not  be  considered.  The  same  will  be  found  to  be  true 
for  the  other  bottom  chords,  so  the  stresses  in  the  bottom  chords  of  this 
bridge,  due  to  wind  —  or  lateral  pressure,  as  the  specifications  call  it  —  can 
be  ignored.  The  same  is  true  for  most  all  ordinary  railroad  bridges. 

178.  Designing  of  the  Top  Lateral  System.  —  The  top  lateral 
system  is  really  a  horizontal  truss  in  the  plane  of  the  top  chords.  In  the 
case  of  through  bridges  (such  as  we  are  designing)  it  is  for  the  purpose 
of  holding  the  top  chords  transversely  and  resisting  the  wind  pressure 
coming  on  the  top  portion  of  the  structure,  but  in  the  case  of  deck  bridges, 
where  the  track  is  on  the  top  of  the  structure,  the  wind  pressure  on  the 
train  and  vibration  due  to  the  train  must  be  provided  for  also,  in  which 
case  the  designing  of  the  top  lateral  system  would  be  the  same  as  shown 
in  the  last  article  for  the  bottom  lateral  system.  The  top  lateral  system, 
as  a  rule,  has  a  double  system  of  diagonals,  the  same  as  the  bottom  lat- 
erals, except  in  the  case  of  through  bridges  a  portal  is  placed  in  each 
end  panel  in  the  plane  of  the  inclined  end  posts  which  takes  the  place 
of  diagonals  in  those  panels.  So,  following  the  usual  practice,  the  top 
lateral  system  in  this  case  will  be  made  as  shown  in  Fig.  269,  where  one 
system  is  dotted  to  avoid  confusion. 


360 


STRUCTURAL  ENGINEERING 


The  lateral  pressure  or  load  which  this  system  must  be  designed 
to  carry,  according  to  the  specifications,  is  200#  per  foot  of  span.  This 
horizontal  load  should  be  considered  as  a  moving  load,  that  is,  as  a  uni- 
form live  load.  Then  for  a  panel  load  we  have 

P  =  25  x  200-  5,000  Ibs., 

one-half  of  which  is  to  be  considered  as  applied  to  each  truss,  that  is, 
wherever  it  is  necessary  to  do  so. 


6  Pane/s  @  2SLO"=.  /SO-  O 


Fig.   269 


Now,  assuming  this  load  as  moving  onto  the  bridge  from  the  right 
end,  and  acting  in  the  direction  indicated  by  the  arrows  (Fig.  269),  we 
obtain 

=  9,250,  say  9,200  Ibs. 
for  the  maximum  tensile  stress  in  diagonal  cD,  and 


10  x 


1.85  =  15,416,  say  15,400  Ibs. 


for  the  maximum  tensile  stress  in  diagonal  bC. 

Assuming  a  full  panel  load  at  each  of  the  points  F  and  E  and  a  half 
panel  load  at  d  (none  at  Z)),  we  have 

5,000      5,000 
3  x  -^- —  +  -~ —  =  5,000  Ibs. 

O  u 

for  the  maximum  compressive  stress  in  strut  dD,  and  assuming  a  full 
panel  load  at  each  of  the  points  F,  E,  and  Z),  and  a  half  panel  load  at  c 
(none  at  C),  we  have 

a     5,000     5,000     „,_,, 
6  x  — - —  +  ~ —  =  7,550  Ibs. 

D  6 

for  the  maximum  compression  in  strut  cC. 

This  really  completes  the  necessary  calculation  for  the  stresses  in 
the  top  lateral  system,  as  will  be  seen  upon  investigation.  For  example, 
when  panel  point  F  alone  is  loaded,  and  the  forces  acting  as  indicated, 
the  stress  in  diagonal  eF  will  be  tension,  as  is  obvious,  but  this  diagonal 
has  maximum  tension  when  the  panel  points  are  loaded  from  the  left 
end  up  to  and  including  point  E  with  forces  acting  in  the  opposite 
direction,  in  which  case  the  stress  in  the  diagonal  would  be  15,400*,  the 
same  as  found  for  diagonal  bC  and  loading  points  F  and  E,  and  the 
forces  acting  as  indicated,  diagonal  dE  would  be  in  tension;  but  the 
maximum  tension  will  occur  in  this  diagonal  when  the  span  is  loaded 


DESIGN  OF  SIMPLE  RAILROAD  BKIDGES  361 

from  the  left  end  up  to  and  including  point  D  and  forces  acting  in  the 
opposite  direction,  in  which  case  the  tension  in  diagonal  dE  would  be 
9,200*,  the  same  as  found  for  diagonal  cD.  So  it  is  seen  that  one  sys- 
tem can  be  considered  in  action  when  the  span  is  loaded  from  one  direc- 
tion and  the  other  system  when  loaded  from  the  opposite  direction,  and 
as  the  lateral  system  is  symmetrical  about  the  center  of  span  it  is  obvious 
that  the  stresses  determined  above  are  sufficient  for  designing  all  the 
members  in  the  system. 

The  members  (diagonals  and  struts)  should  be  as  deep  as  the  top 
chords,  in  order  to  hold  the  top  chords  rigidly.  The  diagonals,  which 
are  usually  designed  as  tension  members,  are  as  a  rule  made  of  two 
angles  latticed  together,  one  angle  being  in  the  plane  of  the  top  of  the 
chord  and  the  other  being  in  the  plane  of  the  bottom  of  the  chord.  The 
struts  (members  cC,  dD,  and  eE)  are  compression  members  and  are,  as 
a  rule,  made  of  four  angles,  two  at  the  top  of  the  chord  and  two  at  the 
bottom.  These  are  latticed  together  in  the  vertical  plane  so  as  to  form 
an  I-section. 

Now,  beginning  with  diagonal  bC  (Fig.  269),  we  have 

15,400 


for  the  required  net  area  of  cross-section  of  the  member,  considered  as  a 
tension  member.  This,  as  is  seen,  (theoretically)  calls  for  two  very 
small  angles,  but  (practically)  the  angles  should  be  such  that  L/r  be  not 
less  than  120,  to  insure  against  vibration.  The  length  of  L  can  be  taken 
as  one-fourth  of  the  total  length  of  the  member. 

(The  diagonals  are  connected  to  each  other  at  their  intersection; 
this  alone  reduces  the  length  one-half,  and  each  half  can  be  considered 
as  a  fixed  column,  thus  reducing  the  length  to  one-fourth  of  the  total 
length.)  The  total  length  of  the  diagonal  center  to  center  of  end  con- 
nections is  about  30  ft.,  or  360  ins.  Then  using  one-fourth  of  this, 
we  have 


for  the  required  radius  of  gyration. 

According  to  this  and  to  the  required  area  of  cross-section,  found 
above,  2£"  x  2J"  or  3"x3"  angles  could  be  used,  but  as  much  better 
details  can  be  obtained  by  using  3J"  x  3J"  angles  —  and  the  weight  of  the 
top  lateral  system  is  a  small  item  —  we  will  use  2  —  Ls  3J"  x  3  J"  x  f  "  for 
each  of  the  diagonals  throughout,  as  cB,  eF,  and  fE  require  the  same 
section  as  bC,  and  the  others,  theoretically,  require  less. 

The  designing  of  the  struts  cC,  dD,  and  eE  is  very  similar  to  the 
designing  of  the  diagonals;  it  is  mostly  a  matter  of  obtaining  rigidity 
and  the  selecting  of  sections  that  are  satisfactory  as  to  details.  But  in 
all  such  cases  the  theoretical  requirement  should  always  be  determined 
before  the  selection  of  section  is  made. 

These  struts  are  more  or  less  fixed  at  their  ends,  but  as  they  are 
wholly  compression  members  we  will  take  the  distance  center  to  center 
of  trusses,  as  their  length,  which  is  16  ft.,  or  192  ins.  In  order  to  insure 


362 


STRUCTURAL  ENGINEERING 


rigidity,  L/r  should  not  exceed  120. 

_193 
r~120 


Then,  taking  L  as  192,  we  have 
=  1.6 


for  the  allowable  maximum  radius  of  gyration.  The  most  satisfactory 
section  for  these  struts  is  four  unequal  leg  angles,  arranged  in  reference 
to  one  another  as  shown  in  Fig.  270.  The  long  legs  of  the  angles  are 
turned  out  so  as  to  make  the  radius  of  gyration  about  axis  y-y  as  large 
as  is  possible,  for,  as  is  obvious,  the  least  radius  is  about  that  axis.  The 
lattice  bars  hold  the  angles  about  J"  apart,  and,  as  the  radius  of  gyration 
about  axis  y-y  is  the  same  for  one  pair  of  angles  as  it  is  for  the  two 
pairs,  we  can  select  the  angles  from  Table  5  by  using  the  angles  having 
r3  equal  to  1.6 — the  required  radius.  As  is  seen  in  this  table,  the  3"  x  2^" 
angles  are  too  small,  as  r3  for  them  is  less  than  1.6;  the  3J"x2J"  are 
large  enough,  but  to  use  a  2|"  leg  would  necessitate  the  using  of  J" 
rivets,  while  J"  would  be  used  in  all  the  other  members;  so  we  will  use 
the  3J"x3"xf"  angles,  in  which  case  r3  is  a  little  more  than  1.71,  as 
seen  in  Table  5.  Now  substituting  this  value  for  r  in  the  column  formula, 
we  have 

1 Q2 
16,000  -  70^7  =  8,140  Ibs.  (about) 

for  the  allowable  unit  stress  on  a  strut  made  of  these  angles.  Dividing 
this  into  the  greatest  stress  found  in  these  struts,  which  is  7,250*,  we 
obtain  less  than  a  square  inch  of  metal  for  the  required  section.  So  it 
is  seen  that  the  stress  does  not  really  influence  the  design  and  that  each 


-f 


± 

Fig.  270 


Fig.  271 


strut  will  be  made  of  4— Ls  3^"x  3"x  f"  =  9.2n",  as  this  fits  the  case 
most  satisfactorily  as  regards  rigidity  and  details. 

179.  Design  of  Portals.— There  are  several  types  of  portals  in 
use.  The  type  shown  in  Figs.  269  and  271  will  be  used  in  this  case,  as 
it  is  a  type  commonly  used;  it  is  quite  simple,  very  rigid,  and  in  fact 
quite  satisfactory  for  ordinary  bridges. 

In  determining  the  stresses  in  portals,  the  first  thing  to  do,  the  same 
as  in  the  case  of  all  frames,  is  to  locate  all  of  the  applied  forces  affecting 
it.  As  is  obvious,  the  maximum  stresses  in  the  portals  will  occur  when 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  363 

all  of  the  panel  points  B  to  F,  inclusive  (Fig.  269),  are  loaded.  As  a 
specific  case,  let  us  consider  the  portal  at  the  left  end  of  the  bridge 
(at  bB)  and  let  us  assume  that  all  of  the  panel  points  are  loaded  and 
that  the  loads  act  as  indicated  in  Fig.  269.  Then  assuming  that  the 
diagonals  carry  only  tension,  the  diagonal  bC  (the  diagonal  connecting 
to  the  portal  at  6)  will  be  in  tension  and  there  will  be  no  stress  in 
diagonal  cB.  One  component  of  the  stress  in  bC  will  be  taken  by  the 
portal  and  the  other  by  the  top  chord  be.  The  component  taken  by  the 
portal  is  applied  (as  is  obvious)  at  b  and  is  equal  to  the  shear  in  panel 
BC,  the  panel  next  to  the  portal.  This  force  we  will  designate  as  R. 
In  addition,  there  will  be  one-half  of  a  panel  load  at  b  and  also  one-half 
of  a  panel  load  at  B.  Designating  the  panel  load  as  P,  we  then  have  the 
loading  on  the  portal'  as  shown  in  Fig.  271:  (.R  +  P/2)at  point  b  and 
P/2  at  point  B.  If  the  pressure,  or  load,  acted  in  the  opposite  direction, 
the  loading  of  the  portal  would  be  just  the  reverse,  that  is,  the  force 
(tf  +  P/2)  would  be  applied  at  B  and  P/2  at  b.  These  forces  applied 
to  the  portal  are  held  horizontally  by  the  two  equal  horizontal  reactions, 
one  at  the  bottom  of  each  end  post.  Let  each  of  these  reactions  be  repre- 
sented  by  //  and  we  have 


These  two  horizontal  reactions  at  the  bottom  of  the  end  posts  and  the 
horizontal  forces,  (.R  +  P/2)  and  P/2,  at  the  top  of  the  portal  form  a 
couple  which  must  be  balanced  by  two  equal  and  opposite  reactions,  V 
applied  along  and  at  the  bottom  of  the  end  posts.  Now,  as  the  forces 
(#  +  P/2),  P/2,  the  two  H's,  and  the  two  F's  balance,  evidently,  the 
frame  formed  by  the  end  posts  and  portal  can  be  treated  as  an  inde- 
pendent structure  in  equilibrium  under  the  action  of  these  forces.  So, 
taking  moments  about  u  (Fig.  271),  and  considering  the  end  posts  and 
portal  combined  as  a  frame,  we  have 


(3). 


or  taking  moments  about  w,  we  have 
- 


In  the  designing  of  portals  for  bridges,  there  are  two  cases:  one 
when  the  end  posts  are  considered  as  hinged  at  the  bottom  ends,  and 
the  other  is  when  the  end  posts  are  considered  as  fixed  at  the  bottom  ends. 
The  top  ends  of  the  end  posts  are  practically  always  considered  to  be  fixed 
by  the  portal.  In  the  case  of  light  bridges,  and  especially  when  end 
floor  beams  are  not  used,  the  end  posts  are  not  very  rigidly  fixed  at  the 
bottom  ends  and  the  actual  condition  is  most  closely  met  by  considering 
the  bottom  ends  of  the  end  posts  hinged,  while  in  the  case  of  heavy 
bridges,  with  end  floor  beams  and  with  wide  shoes,  there  is  no  question 
but  that  the  bottom  ends  of  the  end  posts  are  fixed. 

In  the  case  of  the  bridge  we  are  designing,  there  is  no  question  but 
that  the  end  posts  will  be  quite  rigidly  fixed  at  the  bottom  ends,  but  for 
the  sake  of  illustration  we  will  first  determine  the  stresses  in  the  portals. 


364 


STRUCTURAL  ENGINEERING 


assuming  the  bottom  ends  of  the  end  posts  to  be  hinged,  and  then  deter- 
mine them,  assuming  the  bottom  ends  of  the  end  posts  to  be  fixed. 

By  imagining  the  part  to  the  left  of  section  n-n  (Fig.  271),  including 
the  end  post  bu  and  a  portion  of  the  portal,  moved  bodily  (forces  and  all) 
to  a  new  position  we  obtain  the  structure  shown  in  Fig.  272  which  is 
acted  upon  by  the  known  forces  (R  +  P/2),  H,  -V ,  and  by  the  unknown 
forces  S  and  SI,  S  being  the  stress  in  member  bo  and  SI  the  stress  in  po. 
The  dotted  members  shown  in  the  portal  are  assumed  to  have  no  stress 
in  them  as  they  are  redundant  members  intended  only  to  brace  the  main 
members  of  the  portal,  which  are  shown  in  full  lines,  and  hence  the  dotted 
members  are  ignored  in  the  calculations.  By  taking  moments  about  p 
(Fig.  272)  we  eliminate  £1  and  —V  from  the  equation  of  moments,  and 
as  the  moments  of  H  and  (R  +  P/2)  (about  p)  have  the  same  sign,  both 
tending  to  produce  clock-wise  rotation  about  p,  it  is  evident  that  the  sum 
of  the  moments  of  the  two  forces  about  p  must  be  equal  but  of  opposite 
sign  to  the  moment  of  S  about  p.  So  we  have 


from  which  we  obtain 


for  the  stress  in  portal  member  bo.  Now  it  is  readily  seen  from  Fig.  272 
that  this  stress  S  is  compression,  as  H  and  (R  +  P/2)  both  tend  to  produce 
clock-wise  rotation  about  p,  and  as  S  alone  prevents  this  rotation  it  must 


Fig.  272 


Fig.  273 


evidently  act  to  the  left  toward  b,  and  hence  is  compression.  In  other 
words,  the  member  bo  pushes  to  the  left  against  the  end  post  and  un- 
doubtedly is  in  compression. 

To  determine  the  stress,  SI,  in  the  portal  member  po,  let  us  con- 
sider it  resolved  at  point  p  into  two  components,  one  (/)  along  the  end 
post  and  the  other  (h)  perpendicular  to  the  end  post.  Then  taking 
moments  about  b  (Fig.  272),  we  eliminate  -V,  f,  S,  (R  +  P/2),  and  P/2 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  365 

from  the  equation  of  moments  and  we  have 

from  which  we  obtain 

for  the  component  of  SI  perpendicular  to  the  end  post.     Then  we  have 
SI  =  h  sec  B  =  — 7—  sec  0 (4) 

for  the  stress  in  the  portal  member  (knee  brace)  po.  Stress  S\  is  ten- 
sion, as  is  seen  by  considering  point  b  as  the  center  of  rotation;  the  force 
H  acting  to  the  left  would  necessitate  h  acting  in  the  opposite  direction 
or  away  from  the  post  at  point  p,  which  indicates  the  component  h  to  be 
tension,  and  hence  the  stress  SI  is  tension. 

The  stress  $1  can  also  be  determined  from  the  summation  of  the 
vertical  forces.     In  which  case  we  have 


(-J7  and  /  being  the  only  vertical  forces).     From  this  we  have 

f=r, 

and  multiplying  by  sec  </>  we  obtain 

sec<£  ...................  (5). 


/  equals  V  ,  and  as  these  two  forces  must  balance  each  other  it  is  seen 
that  /  acts  as  indicated  in  Fig.  272,  which  shows  $1  to  be  tension. 

By  imagining  the  part  to  the  right  of  section  n'-nf  (Fig.  271),  in- 
cluding end  post  wB  and  a  portion  of  the  portal,  moved  bodily  (forces 
and  all)  to  a  new  position  we  obtain  the  independent  structure  shown  in 
Fig.  273  acted  upon  by  the  known  forces  H,  V  ,  and  P/2  and  by  the 
unknown  forces  (stresses)  S2  and  S3.  Taking  t  as  the  center  of  moments, 
it  is  seen  that  P/2  and  H  both  tend  to  produce  clock-wise  rotation,  and  as 
this  rotation  is  prevented  by  S2  alone,  we  have 


from  which  we  obtain 
P     Hm 


for  the  stress  in  portal  member  Bo.  This  stress  is  tension,  as  is  readily 
seen,  as  it  acts  away  from  the  end  post  at  point  B,  that  is,  the  member 
Bo  pulls  on  the  end  post  at  point  B  and  hence  is  in  tension. 

Resolving  (at  point  t)  the  stress  S3  into  two  components  h'  and  /', 
nnd  taking  moments  about  B,  we  have 


366  STRUCTURAL  ENGINEERING 

from  which  we  obtain 


and  multiplying  by  sec  0  we  have 

S3  =  h'  sec  0  =  ^-^sec  0  ..............................  (7) 

for  the  stress  in  portal  member  ot. 

By  considering  point  B  (Fig.  273)  as  the  center  of  moments,  the 
stress  S3  is  seen  to  be  compressive,  as  //  acting  to  the  left  requires  the 
member  ot  to  act  to  the  right  against  the  end  post,  and  hence  the  member 
ot  is  in  compression. 

As  is  seen  by  comparison,  (4)  and  (7)  are  exactly  alike.  That 
means  that  the  stress  in  po  is  equal  to  the  stress  in  ot,  but,  as  stated 
above,  one  is  tension  and  the  other  is  compression. 

Now,  having  derived  a  formula  for  the  stress  in  each  member  of  the 
portal,  we  will  proceed  with  the  determination  of  the  stress  in  the  mem- 
bers. To  do  this  it  is  first  necessary  to  determine  the  value  of  P,  R,  H, 
L,  m,  k,  and  sec  0.  P  is  given  in  the  last  article  as  5,000*.  R  is  equal 
to  the  shear  in  the  panel  BC  (Fig.  269)  when  all  the  panels  are  loaded. 
Beginning  at  the  center  of  the  span,  we  have  one-half  of  a  panel  load  at 
D  and  a  full  panel  load  at  C,  making  in  all  one  and  one-half  panel  loads 
as  the  shear  in  the  panel  BC,  hence  we  have 

R  =  ^9.  +  5,000  =  7,500  Ibs. 
0 

for  this  shear.     Then  from  (1)  we  have 

H=  7,500  +  5,000  =6)2501bs- 
4 

L  is  about  39  ft.,  as  the  height  of  the  span  is  30  ft.  and  the  panel  length 
is  25  ft.  It  will  be  close  enough,  for  designing,  to  take  B  as  45  degrees. 
Then,  A:  =  8  ft.,  m  =  31,  and  sec  0=1.4.  Now,  using  the  above  values, 
from  (3)  we  obtain 

£  =  7,500+^^  +  6,250x^  =  34,218,  say  34,000  Ibs. 

A  O 

(compression)  for  the  stress  in  portal  member  bo.  From  (4)  or  (7) 
we  obtain 

Sl  =  S3=  6^X39  x  1.4  =  42,655,  say  43,000  Ibs. 
o 

for  the  tensile  stress  in  portal  member  po  and  compressive  stress  in  ot. 
From  (6)  we  obtain 

6,850x31 


j_ 

a  O 

for  the  tensile  stress  in  the  portal  member  Bo.  By  comparing  (3)  and 
(6)  it  is  seen  that  the  stress  in  member  Bo  differs  from  the  stress  in 
member  bo  bv  the  value  of  R. 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES 


367 


h 


(OJ 


F 


(b) 


This  completes  the  calculations  for  the  stresses  in  the  portals  under 
the  assumption  that  the  bottom  ends  of  the  end  posts  are  hinged,  and 
we  will  now  determine  the  stress  in  the  portal,  assuming  the  bottom  ends 
of  the  end  posts  fixed.  To  illustrate  conditions,  let  e  and  g  at  (a)  (Fig. 
274)  represent  two  thin  wooden  strips  connected  rigidly  by  another  piece  h. 
If  this  frame  be  supported  as  shown  at  (6)  and  a  force  F  be  applied, 
the  pieces  e  and  g  would  bend  as  indicated  at  (6),  and  we  thus  have 
an  illustration  of  the  case  just  considered 
where  the  end  posts  were  assumed  hinged  at 
the  bottom  ends.  Now,  suppose  the  bottom 
ends  of  the  pieces  to  be  buried  in  concrete  as 
indicated  at  (c)  and  the  concrete  be  allowed 
to  set  and  the  force  F  be  then  applied.  The 
pieces  e  and  g  would  then  bend  as  indicated  at 
(c),  and  we  thus  have  an  illustration  of  the 
case  to  be  considered  where  the  bottom  ends 
of  the  end  posts  are  assumed  to  be  fixed.  The 
points  o'  are  the  points  of  contra-flexure,  where 
the  bending  moment  on  the  pieces  e  and  g  is 
zero,  as  is  readily  seen  from  the  sketch.  The 
points  of  are  at  mid-distance  from  the  con- 
crete to  the  piece  h,  or  m/2  from  h,  neglecting 
the  portion  buried  in  the  concrete  as  being  too 
small  to  materially  change  the  length  of  the 
part  marked  m. 

Now,  as  there  is  no  bending  at  the  points 
o'  (points  of  contra-flexure),  the  frame  can  be 
assumed  as  cut  off  at  those  points  and  we  have 
the  independent  structure  shown  at  (d).  From 
this  it  is  seen  that  the  problem  of  determin- 
ing the  stresses  in  portals,  when  the  bottom 
ends  of  the  end  posts  are  considered  fixed,  is 
the  same  as  in  the  case  of  hinged  ends,  except 
the  distance  m  (Figs.  271  and  273)  is  one- 
half  as  great,  and  hence  the  above  formulas  for 
determining  the  stresses  in  the  portals  are  ap- 
plicable if  m/2  be  substituted  for  m  and 
(L-m/2)  or  (&  +  ra/2)  for  L.  In  an  imaginary  sense,  we  simply  move 
the  bottom  supports  of  the  end  posts  half  way  up  their  unsupported 
length,  and  proceed  with  the  determination  of  the  stress  in  the  portals  in 
the  same  manner  as  if  the  bottom  ends  of  the  end  posts  were  hinged. 
Thus  substituting  in  (3),  we  have 

£  =  7,500  +  M^1  +  6,250x^  =  22,109,  say  22,000  Ibs.  (compression) 


F 


'V 
^ 
^ 

-^=-=3. 

o' 

0' 

\\ 

\\\ 

\N\\\\\\\\\V 

VW1 

(C) 


Fig.   274 


for  the  stress  in  portal  member  bo  (Fig.  271),  assuming  the  bottom  ends 
of  the  end  posts  fixed;  and  substituting  in  (4),  we  have 

6180  *  (30  -V) 

8 


SI  = 


=  25,703,  say  26,000  Ibs. 


for  the  tensile  stress  in  portal  member  po  and  also  for  the  compressive 


368  STRUCTURAL  ENGINEERING 

stress  in  portal  member  oiy  and  likewise  substituting  in  (6)  we  have 

S2  =  P    *-£  a  Mgi  +  m;  -  V  =  14,609,  say  15,ooo  lbs. 

6  K  £  O 

for  the  tensile  stress  in  portal  member  Bo.  Thus  we  have  all  the  stresses 
in  the  portals  determined  when  assuming  the  bottom  ends  of  the  end  posts 
to  be  fixed.  These  stresses  will  be  used  in  designing  the  sections  of  the 
portal  members  as  the  bridge  considered  is  designed  for  a  heavy  loading; 
and  we  have  provided  end  floor  beams,  consequently  the  bottom  ends  of 
the  end  posts  will  be  fixed  quite  securely. 

Each  member  of  the  portals  will  be  as  deep  as  the  end  posts  and 
composed  of  two  angles,  latticed  together  vertically,  similar  to  the  top 
laterals. 

As  seen  above,  the  members  are  subjected  to  one  kind  of  stress 
when  the  wind  load  comes  from  one  direction  and  just  the  reverse  when 
it  comes  from  the  opposite  direction,  that  is,  each  portal  member  is  sub- 
jected to  reversal  of  stress,  and  hence  the  members  must  be  designed  to 
carry  both  compression  and  tension. 

The  members  po  and  ot  (known  as  knee  braces)  are  each  subjected 
to  26,000*  stress,  considered  to  be  either  tension  or  compression.  Taking 
the  case  of  tension  first,  we  have 

26,000 


for  the  required  net  area  of  cross-section  of  each,  which  shows  that  two 
very  small  angles  could  be  used  (theoretically)  as  far  as  tension  is  con- 
cerned. As  for  compression,  the  length  of  each  member  is  about  11.3  ft. 
(0  =  45°),  but  as  the  redundant  members  (those  dotted,  in  Fig.  271)  sup- 
port each  at  mid-point,  the  length  L  can  be  taken  as  11.3/2  =  5.6  ft.,  say 
66  ins.  As  the  members  are  subjected  to  reversal  of  stress,  L/r  should 
be  reasonably  low,  say  80.  Then  we  have 


, 

for  the  required  value  of  the  radius  of  gyration. 

Now,  glancing  over  the  table  of  angles  (Tables  5  and  6)  it  is  seen 
that  this  calls  for  very  small  angles.  So  the  case  here  is  the  same  as 
that  of  the  top  lateral  struts  —  very  much  a  matter  of  judgment.  3£"  x 
3J"xf"  angles  are  as  small  as  are  used  in  this  class  of  work  (as  it  is 
desirable  to  use  J"  rivets  throughout),  so  we  will  try  that  section.  Sub- 
stituting the  radius,  found  in  Table  6,  for  this  angle  in  the  column 
formula,  we  have 

16,000-70^  =11,682  lbs. 

for  the  allowable  unit  compressive  stress  in  each  of  the  members  po  and 
ot.  Dividing  this  into  the  stress,  we  have 

26,000  -r  11,682  =  2.22  sq.  ins. 

for  the  required  area,  while  2—  Ls  3£"  x  3£"  x  f  "  have  4.96n",  more  than 
twice  the  amount  required  for  compression,  and,  as  found  above,  only 
1.62n"  is  required  for  tension.  So  it  is  seen  that  3£"  x  3£"  x  f  "  angle/? 


DESIGN  OF  SIMPLE  RAILEOAD  BKIDGES 


369 


are  much  heavier  than  required,  theoretically,  yet  smaller  angles  would 
give  the  portals  a  flimsy  appearance  even  if  satisfactory  details  were 
obtained  by  their  use.  So  we  are  now  to  the  point  where  the  selection 
must  be  made  mostly  upon  appearance.  As  the  stress  in  the  members 
bo  and  Bo  is  less  than  in  members  po  and  ot,  and  their  lengths  are  also 
less,  it  is  not  necessary  to  compute  the  required  sections  for  those  mem- 
bers, but  use  the  same  as  used  for  members  po  and  ot.  As  the  3J"  x  3  J" 
angles  are  our  minimum  size  we  could  use  two  of  these  for  each  portal 
member,  but  for  the  sake  of  details  and  appearance  we  will  use  2 — Ls 
4"x4"xf"  for  each  of  the  main  members  and  2— Ls  3J"  x  3$"  x  f"  for 
each  of  the  redundant  members,  thus  giving  the  main  members  a  dis- 
tinctive outline. 

In  the  case  of  large  bridges  these  members  can  be  designed  theoretic- 
ally correct,  as  the  required  area  of  cross-section  in  those  cases  is  such 
that  sections  can  be  found  which  are  both  theoretically  and  practically 
correct,  but  in  the  case  of  small  bridges  it  is  very  much  a  matter  of 
judgment,  as  is  seen  from  the  above  calculations. 

180.  Design  of  End  Posts. — The  designing  of  the  section  of  the 
end  posts  is  here  taken  as  a  special  problem  for  the  reason  that,  unlike 
the  other  main  truss  members,  it  is  subjected  to  bending  stress  in  addi- 
tion to  direct  stress.  To  correspond  with  the  top  chord,  it  will  be  made 
of  18"  webs,  23"  cover  plate  and  four  angles.  The  length  of  the  end 


post,  considering  the  x-x  axis  (see  Fig.  265)  is  the  total  length  L  (shown 
in  Fig.  275  to  equal  AB)  unless  a  collision  strut  OM  be  inserted,  in  which 
case  the  length  is  L/2  and  the  length,  considering  axis  y-y,  is  equal  to  m. 
The  collision  strut  should  be  used  as  a  safeguard  in  case  a  derailed  train 
should  strike  the  end  post,  and  therefore  we  will  use  one  in  this  case. 
Then  the  length  of  the  post  in  reference  to  the  x-x  axis  is  1  9.5  ft.  or  234 
ins.,  and  31  ft.  (=ra)  or  372  ins.  in  reference  to  axis  y-y. 

Using  the  radii  found  for  top  chord  be   (Art.  176),  which  will  be 
about  the  same  in  this  case,  we  obtain 


. 
16,000  -  70  ^-  =  1  3,750  Ibs. 

i  .^O 

for  the  allowable  unit  direct  compressive  stress  in  the  end  post,  consider- 

ing axis  x-x,  and 

379 
16,000-  70         =  12,678  Ibs. 


370  STRUCTURAL  ENGINEERING 

for   the   allowable   unit   direct   compressive   stress   considering   axis   y-y. 
Now  dividing  the  latter,  as  it  is  the  smaller,  into  the  totnl  direct  stress 
in  the  end  post,  as  given  in  Fig.  262,  we  obtain 
"  523,000  . 


for  the  required  area  of  cross-section  of  each  end  post,  providing  there 
were  no  cross  bending,  due  to  the  wind  load,  on  the  posts.  But  as  the 
bending  stress  on  the  compression  side  of  the  posts  adds  to  the  direct 
stress,  the  two  unit  stresses  must  be  combined.  However,  in  case  these 
stresses  be  combined,  the  allowable  unit  stress  can  be  increased  25  per 
cent,  so  the  above  area  may  be  about  correct.  As  preliminary  let  us  try 
the  section  used  for  top  chord  be,  which  contains  40.5Sn". 

In  determining  the  bending  moment  on  the  end  posts,  which  is  due 
to  wind  —  and  consequently  a  transverse  moment  —  we  can  consider  each 
post  as  composed  of  two  cantilevers  as  indicated  at  (a)  in  Fig.  275. 
Then  the  moment  on  each  post  at  any  point  either  above  or  below  the 
point  of  contra-flexure  (o')  is  Hx.  Now  it  is  seen  that  this  moment  is  a 
maximum  at  the  points  p  and  t  and  at  the  bottom  of  each  post,  and  that 
the  maximum  in  each  case  is  equal  to  Hm/2.  As  the  points  p  and  t  are 
the  farthest  out  from  the  ends  of  the  member,  where  the  column  formula 
really  applies,  we  will  consider  the  bending  at  those  points  only.  From 
the  last  article,  we  have  H  =  6,250  Ibs.  and  m  =  31ft.  or  372  ins.  Then 
for  the  moment  at  p  or  t  (Fig.  275  (a))  we  have 

M  =  6,250  X~^=  1,162,500  inch  Ibs. 

a 

Now  using  the  moment  of  inertia  found  in  Art.  176  for  chord  be  in 
reference  to  axis  y-y  and  taking  the  horizontal  distance  to  the  extreme 
fiber  as  7.75  +  4=11.75"  (see  Fig.  265),  we  obtain 


for  the  unit  stress  due  to  cross  bending,  assuming  that  the  same  section 
as  used  for  chord  be  is  used  for  the  end  post. 

Now  dividing  the  area  of  the  section  into  the  total  direct  stress, 
we  have 

.  (about)  :'- 


for  the  actual  direct  unit  stress,  and  adding  this  to  the  unit  stress  due 
to  cross  bending  we  have 

5,470  +  12,880  =  18,350  Ibs. 

for  the  combined  unit  stress.  The  allowable  unit  stress  as  given  above  is 
12,678.  But  this  can  be  increased  25  per  cent  in  the  case  of  combined 
stress,  as  per  specifications.  So  we  have 

(12,678)  1.25  =  15,847  Ibs. 

for  the  actual  allowable  combined  unit  stress,  and  as  it  is  less  than  the 
above  (18,350#)  a  larger  section  will  have  to  be  used  for  each  end  post 
than  was  just  considered. 


DESIGN  OF  SIMPLE  EAILEOAD  BKIDGES 


371 


Let  us  assume  the  following  sections: 

1— cov.  pi.  23"x|"  =  11.50n" 
2 — web  pis.  18"  x4"  =  18.00°" 
2— Ls  3J"  x  3V  x  |"  =  4.96n// 
2— Ls  6"x4"xf"  =11.72n" 

46.18°" 

As  the  post  is  stronger  in  reference  to  the  x-x  axis  than  in  reference  to 
the  y-y  axis,  owing  to  the  length  being  greater  in  the  last  case  (using 
the  collision  strut),  we  need  consider  the  above  section  only  in  reference 
to  the  y-y  axis. 

O     J 

For  the  moment  of  inertia  about  the  y-y  axis  (see  Fig.  276),  we  have 
cov.  pi.  507  +  0  =507  (about) 

web  pis.  0  +  7^622  x  9  x  2  =1,045 

top  angles  2.87  x  2  +  £882x  2.48  x  2  =     397 
bot.  angles  7.52  x  2  +  SlKJ^x  5.86  x  2  =     943 

2,892 
Then  for  the  radius  of  gyration  about  axis  y-y  we  have 


Now  substituting  this  radius  in  the  column  formula,  we  have 
p  =  16,000  -  70  JEU  12,710  (about) 


for  the  allowable  unit  stress  for  direct  compression, 
and  increasing  this  25  per  cent  we  obtain  15,890* 
for  the  allowable  unit  stress  in  the  case  of  direct 
compression  and  bending  stress  combined.  Then, 
using  the  last  assumed  section,  we  have 


11,320  Ib,  (about) 


for  the  actual  direct  unit  compressive  stress,  and 

1,162,500x11.75 
- 


Fig.  276 


for  the  unit  stress  due  to  bending.  Now,  adding  these  two  stresses 
together,  we  have  11,320  +  4,730  =  16,050*  for  the  combined  unit  stress, 
which  is  only  160*  (=16,050-15,890)  greater  than  the  15,890*  allowed, 
and  as  this  is  about  as  close  as  we  can  obtain  we  will  use  the  above 
section  for  each  end  post. 

181.  Designing  of  Collision  Struts.  —  These  members  should  at 
least  have  sufficient  section  to  resist  the  thrust  that  an  ordinary  loaded 
car  moving  at  an  ordinary  rate  would  exert  upon  them  in  case  the  car 
were  derailed.  The  problem  involved  is  of  such  a  nature  that  the  entire 


372 


STRUCTURAL  ENGINEERING 


premises  must  rest  upon  practical  assumptions.  We  will  assume  that  the 
car  and  its  load  weigh  100,000*  and  that  it  has  a  velocity  of  30  ft.  per 
second,  which  is  a  little  over  20  miles  per  hour,  and,  further,  we  will 
assume  the  car  brought  to  rest  in  one  second.  Then  substituting  in 
Formula  A  (Art.  23),  we  have 

=93,750,  say  94,000  Ibs., 


for  the  constant  force  required  to  stop  the  car  in  one  second.  But  at 
the  beginning  of  contact  the  force  would  be  zero  and  a  maximum  when  the 
car  was  brought  to  rest,  so  the  maximum  force  exerted  by  the  car  would 
really  be  twice  the  constant  force  required  to  stop  it,  or  94,000x2  = 
188,000*.  This  force  would  be  exerted  on  the  end  post  at  a  point  about 
9.5  ft.  above  the  bottom  chord  (see  Fig.  277),  so  that  the  horizontal  force 
exerted  against  the  collision  strut  would  be  about  9.5/15  of  the  188,000, 
or  119,000*. 

Then  multiplying  one-half  of  this  by  sec  /?,  which  we  will  take  as 
1.56,  (3  being  the  complement  of  0,  we  have 


119,000 


Xlt56  =  92^820,  say  93,000  Ibs. 


for  the  stress  in  the  collision  strut.     It  may  appear  that  the  collision  strut 
should  connect  to  the  end  post  at  the  point  where  the  188,000*  is  applied, 

but  this  is  not  the  case,  for  if  the  blow 
of  188,000*  were  struck  at  the  point 
where  the  strut  connects  to  the  end 
post  there  would  be  but  little  resilience 
and,  consequently,  the  structure  would 
be  subjected  to  severe  shock. 

It  is  seen  from  the  amount  of  stress 
that  the  section  of  the  collision  struts 
need  not  be  large  as  far  as  direct  stress 
is  concerned,  but  the  value  of  L/r  should 
not  exceed  80  in  order  to  obtain  rigidity. 
The  best  section  for  these  struts  is  two  channels  latticed  together,  and 
as  8"  channels  are  about  the  smallest  used  in  this  class  of  work  we  will 
try  2  —  [s  8"  x  16.25*.  The  length  of  each  strut,  center  to  center  of 
end  connections,  is  about  19.5  ft.,  or  234  ins.,  and,  as  seen  from  Table 
3,  the  radius  of  gyration  of  the  channels  is  2.89,  so  we  have 


Fig.  277 


234 

2.89 


=  81. 


Then  for  the  allowable  unit  compressive  stress  on  each  strut  we  have 

p  =  16,000  -  70  x  81  =  10,330  Ibs. 
Now,  dividing  this  into  the  stress,  we  have 

93,000 


for  the  section  required.     The  area  of  the  section  of  the  two  channels 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


373 


assumed,  as  seen  from  Table  3,  is  9.56n",  which  is  0.56n"  more  than  the 
required  area,,  but  we  will  use  this  section  as  it  is  about  as  close  to  the 
required  area  as  we  can  come  and  at  the  same  time  obtain  good  details. 

The  above  analysis  is  only  an  attempt  to  provide  for  reasonable 
accidents,  as  we  might  term  it.  The  actual  intensity  of  the  blow  struck 
by  a  derailed  car  is  distinctively  problematic  for  the  reason  that  there 
could  be  so  many  different  conditions  assumed.  As  an  illustration,  sup- 
pose a  derailed  car  in  the  middle  of  a  train  should  strike  the  end  post 
of  a  bridge.  How  much  would  the  cars  in  the  rear  of  the  derailed  car 
increase  the  blow?  And  how  much  would  the  engine  pulling  on  the 
train  increase  it?  And  how  much  would  the  resistance  of  the  wheels  of 
the  derailed  car  bumping  over  the  ties  diminish  it?  And,  further,  sup- 
pose a  train  traveling  50  miles  per  hour  should  jump  the  track  just  as 
it  gets  to  the  bridge  and  the  engine  should  strike  the  end  post.  There  is 
little  doubt  but  that  the  bridge  would  be  destroyed,  yet  it  might  glance 
off  and  do  but  little  damage.  It  is  obvious  that  it  would  be  absolutely 
impracticable  to  provide  for  resisting  the  blow  in  the  last  case ;  and  prac- 
tically the  same  is  true  in  several  cases  that  could  be  assumed,  yet  it 
would  not  be  good  engineering  to  make  no  provision  for  reasonable  acci- 
dents, as  past  experience  has  taught  us,  and  that  is  what  we  have 
endeavored  to  do  above. 

182.  Maximum  Reaction  on  Shoe. — For  the  dead-load  reaction  on 
each  shoe  we  have  one-half  of  a  panel  load  from  D  (Fig.  249),  a  full 
panel  load  from  B  and  C  each,  and  one-half  (so  considered)  of  a  panel 
load  from  A,  making  in  all  three  panel  loads  on  each  shoe.  Then  taking 
the  dead-load  panel  load  as  found  in  Art.  173,  we  have 

12-3x26,375  =  79,125,  say  80,000  Ibs. 

for  the  dead-load  reaction  on  each  shoe. 

As  is  evident,  the  maximum  live-load  reaction  on  a  shoe  will  occur 
when  the  span  is  fully  loaded  and  the  heaviest  wheels  are  near  the  shoe 


Fig.   278 


and  one  directly  over  it — that  is,  over  the  end  floor  beam.  So,  placing 
the  live  load  as  shown  in  Fig.  278  with  wheel  (2)  at  A  and  taking 
moments  about  G,  using  Table  A,  we  have 

15,274  +  274  x  49  +  49^  50 


R'  = 


150 


for  the  maximum  live-load  reaction  on  the  shoe  at  A,  which  is  the  same 
as  for  the  others,  and,  for  the  impact,  we  have 


259,000  x 


,  say  173,000  Ibs. 


1J: 


:•" 


I: 

ilf*3S    J  £33  §>  ^ 

-!-i  1^  1^^?^  .v 

l.S  l       ^     nrd3  I? 


374 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


375 


Now  adding  the  above  dead,  live,  and  impact  reactions  together,  we  have 

80,000  +  259,000  +  173,000  =  512,000  Ibs. 
for  the  maximum  reaction  on  each  shoe. 

183.  Stress  Sheet. — After    having   completed    the    calculations    as 
above  for  the  span  the  "stress  sheet,"  Fig.  279,  can  be  made.     This  draw- 
ing contains  all  the  information  resulting  from  the  calculations;  in  fact, 
the  drawing  is  really  a  summary  of  the  calculations.     After  the  stress 
sheet  (Fig.  279)  is  completed  the  detail  drawings  can  be  made. 

184.  Calculations  of  Details. — In  starting  the  detail  drawings,  a 
large  scale  pencil  sketch  (1J",  2"  or  3"  scale)  of  each  joint  should  be 
made  first.     Such  a  drawing  for  joint  LO  is  shown  in  Fig.  282.     The 
calculations  of  the  details  are  made  as  the  sketches  are  being  drawn. 

Joint  LO.  Taking  first  the  case  of  joint  LO  (Fig.  282),  the  first 
thing  to  do  is  to  locate  the  center  line  of  the  pin  in  the  end  post.  The 
section  of  the  end  post  is  given  on  the  stress  sheet  (Fig.  279),  also  in 
Art.  180. 

Taking  moments  about  the  center  of  the  cover  plate  of  the  section 
(see  Fig.  280)  we  have 

-     1.26  x  4.96  + 18  x  9.37  -f  11.72  x  16.47     '367.9 


x  = 


46.18 


46.18 


=  7.96  ins. 


for  the  distance  from  the  center  of  the  cover  plate  down  to  the  gravity  axis 
x-x  of  the  end  post.  Now,  as  far  as  the  direct  stress  in  the  end  post  is 
concerned,  the  pin  could  be  properly  placed  on  this  gravity  axis  x-x;  but 
the  member  tends  to  bend  downward  as  shown  in  Fig.  281,  owing  to  its 
own  weight,  and  it  is  necessary  to  place  the  pin  a  short  distance  below 
the  gravity  axis  so  as  to  counterbalance  this  bending.  The  pin  being 
placed  below  the  gravity  axis  x-x,  the  direct  stress  tends  to  bend  the 
member  upward  as  indicated  by  the  dotted  line,  and,  as  is  evident,  if  the 
pin  be  placed  just  the  correct  distance  below  the  gravity  axis  the  end  post 
will  be  straight  when  the  maximum  stress  in  it  occurs.  For  the  weight  of 
the  end  post  per  foot  of  length  we  have 


a 

NJ 

v..^  -t 

If 

" 

0) 

'* 

^VVV.V'A              , 

Fig.   28U 


Fig.   281 


W  =  46.18n"  x  3.4#  =  157  Ibs., 

and  increasing  this  one-third  to  provide  for  details  we  have  210*  for  the 
total  weight  of  the  member  per  foot  of  length.  This  weight  acts  vertically 
and,  as  the  member  is  inclined,  only  the  component  perpendicular  to  the 
member  causes  transverse  bending.  So,  for  the  bending  moment,  we  have 


Fig.   282 


376 


DESIGN  OF  SIMPLE  KAILEOAD  BRIDGES  377 

M  =  J  x  210  x  sin<9  x  19J?  x  12  =  76,658  inch  Ibs.     (Sin0  =  0.64.) 
Now  by  placing  the  pin  a  distance  z  below  the  gravity  axis  x-x,  we 
have  2  x  523,000  for  the  moment  caused  by  the  direct  stress  which  tends 
to  bend  the  end  post  upward,  and  hence  the  end  post  will  be  straight 
when  2  x  523,000  =  76,658,  from  which  we  obtain 


. 

which  is  the  distance  that  the  pin  should  be  placed  below  the  gravity  axis 
x-x.  Now  adding  this  to  izT  (given  above)  we  have  7.96  +  0.1466  =  8.1066" 
for  the  distance  from  the  center  of  the  cover  plate  down  to  the  center  of 
the  pin,  and  substracting  one-half  of  the  thickness  of  the  cover  plate  from 
this  we  have  8.1066  -  0.25  =  7.8566"  for  the  distance  from  the  under  side 
of  the  cover  plate  to  the  pin.  The  distance  from  the  under  side  of  the 
cover  plate  to  the  working  line  in  the  top  chord,  which  would  be  the  cen- 
ter line  of  the  pins  if  the  top  chord  were  pin-connected,  should  be  the 
same  as  for  the  end  post  in  order  to  obtain  uniform  construction  of  these 
members,  and  hence  at  this  juncture  it  is  necessary  to  examine  the  top 
chord  sections  and  select  a  distance  suitable  for  all  top  chord  members 
and  end  posts. 

Considering  the  case  of  the  top  chord  section  U2-U3  (Fig.  279),  and 
taking  moments  about  the  cover  plate  as  was  shown  above  in  the  case  of 
the  end  post  we  obtain 

-     1.26  x  4.96  +  18  x  9.37  +  10.62  x  16.49 
'  =  —•  -  451)8  --  =7'76  ^ 

for  the  distance  from  the  center  of  the  cover  plate  down  to  the  horizontal 
gravity  axis.  For  the  weight  of  the  section  we  have 

45.08  x  3.4  =  153  Ibs. 

per  foot  and  adding  one-third  for  details  we  have  204#  for  the  weight  of 
the  member  per  foot  of  length.  Then  we  have 

M  =  J  x  204  x  252  x  12  =  191,000  Ibs.   (about) 

for  the  bending  on  the  chord  due  to  its  own  weight,  and  for  the  distance 
z'  below  the  horizontal  gravity  axis,  where  pins  would  have  to  be  placed 
to  balance  the  above  moment,  we  have 

191,000 


Adding  this  to  the  above  7.76"  and  subtracting  one-half  of  the  thickness 
of  the  cover  plate  we  have 

7.76  +  0.325  -  0.25  =  7.84  ins. 

for  the  distance  down  from  the  under  side  of  the  cover  plate  to  the  work- 
ing line.  In  the  same  manner  we  find  that  the  corresponding  distance  in 
the  case  of  chord  £71-172  is  7.75".  So  it  is  seen  that  7£"  is  about  the  cor- 
rect value  to  use  for  the  distance  from  the  under  side  of  the  cover  plate  to 
the  pin  center  or  working  line,  as  the  case  may  be,  and  hence  this  distance 
will  be  taken  in  the  case  of  all  top  chord  sections  and  end  posts  throughout 
the  structure.  Thus  we  have  the  pin  LO  located  and  the  outline  of  the 
portion  of  the  end  post  shown  (Fig.  282)  can  be  drawn  and  the  rivets 


378  STRUCTURAL  ENGINEERING 

spaced  transversely  in  it  as  shown  and  then  the  drawing  of  the  other 
details  can  proceed.  The  end  post  should  extend  far  enough  beyond  the 
pin  so  that  the  bottom  chord  and  end  floor  beam  connections  balance  fairly 
well  about  the  pin.  This  part  is  at  first  determined  merely  by  appearance. 
After  this  the  shoe  can  be  sketched  so  as  to  clear  the  end  post,  the  number 
of  rollers  required  is  determined  by  sketching  roughly  their  length,  and  at 
the  same  time  the  length  of  the  pedestal  can  be  determined.  After  this  is 
done  the  required  pin  bearing  on  the  end  post  can  be  figured.  This  bear- 
ing must  be  sufficient  to  carry  the  maximum  vertical  reaction,  which  is 
512,000*  (see  stress  sheet,  Fig.  279).  Let  us  assume  a  5£"  pin.  Then 
we  have 

t  =  512,000  -r  (5J  x  24,000)  =  3.87  ins. 

for  the  required  thickness  of  bearing  on  the  two  sides  of  the  end  post  or 
1.93"  (=  1-Jf )  for  each  side.  As  shown,  we  have  a  f "  gusset  plate,  \" 
web,  f "  filler,  and  a  TV'  plate,  making  in  all,  f  +  i  +  f  +  iV  =  ^iV*  which 
is  \"  thicker  than  required.  But  this  is  about  as  close  as  we  can  come, 
as  the  gusset  plate  should  be  f "  thick  (as  will  be  seen  later)  and  the  f" 
filler  can  not  be  reduced,  as  the  bottom  angles  on  the  end  post  have  that 
thickness  and  the  iV  plate  is  as  thin  as  should  be  used,  owing  to  the  coun- 
tersunk rivets  (  J"  rivets  should  not  be  countersunk'  in  metal  less  than 
•f$"  thick).  So  we  will  use  the  metal  shown.  There  should  be  enough 
rivets  above  the  pin  to  hold  these  plates.  First,  there  should  be  enough 
rivets  connecting  the  f "  filler  and  the  •£$"  plate  to  the  end  post  and  gus- 
set plate  to  transmit  the  pin  pressure  coming  on  the  two.  For  this  pres- 
sure we  have  (f  +  T^)  5J  x  24,000  =  140,000*  (about).  The  rivets  are  f" 
shop  rivets  in  single  shear,  each  of  which  is  good  for  12,000  x  0.6  =  7,200*. 
So  we  have  140,000 -=-7,200  =  19.5,  say  20  rivets.  We  have  just  about 
20  rivets  above  the  pin,  so  the  detail  is  satisfactory  so  far.  The  rivets  in 
the  y  filler  are  not  included,  as  they  simply  hold  that  narrow  filler. 
There  should  be  enough  rivets  in  the  -fa"  plate  to  take  the  pin  pressure 
upon  it,  which  is  &  x  5J  x  24,000  =  57,750*.  Dividing  this  by  7,200  we 
have  57,750  -r  7,200  =  8  rivets — whereas  we  have  over  twice  that  number, 
and  that  part  is  amply  strong.  To  be  on  the  safe  side  there  should 
be  enough  rivets  connecting  the  end  post  to  the  gusset  plate  to  trans- 
mit, in  single  shear,  one-half  of  the  stress  in  the  end  post,  which  is 
523,000 -=- 2  =  261,500  (see  Fig.  279).  Dividing  this  by  7,200  we  have 
261,500 -=-7,200  =  36.3,  say  37  rivets.  We  have  about  38  above  the  pin, 
so  that  part  of  the  detail  is  satisfactory.  We  will  next  consider  the  bot- 
tom chord  connection.  The  net  area  of  the  bottom  chord  is  20. 9n"  (see 
Fig.  279),  and  for  its  strength  we  have  20.9  x  16,000  =  334,400*.  Then 
as  the  rivets  are  I"  field  rivets,  we  have  334,400  -=-  6,000  =  55.7,  say,  56 
rivets,  or  28  on  each  side.  We  have  30,  so  that  detail  is  correct. 

The  end  floor  beam  is  cut,  as  shown  (Fig.  282),  to  clear  the  shoe,  pin, 
and  the  details  on  the  end  post.  Dividing  the  end  shear  on  the  end  floor 
beam  (see  Fig.  279)  by  6,000  we  have  143,200 -=-6,000  =  23.8,  say  24 
rivets  for  the  number  of  J"  field  rivets  required  in  each  end  connection. 
As  is  seen,  24  rivets  are  used.  The  thickness  of  pin  bearing  on  the  shoe 
need  be  (theoretically)  but  1-Jf "  on  a  side,  as  found  above  for  the  end 
post,  but  there  should  be  a  point  of  bearing  on  each  side  of  each  web  of 
the  end  post  (as  shown)  so  as  to  distribute  the  pressure  well  over  the 


DESIGN  OF  SIMPLE  KAILEOAD  BRIDGES  379 

rollers  and  at  the  same  time  lighten  the  stress  on  the  pin,  and  as  every 
part  of  such  castings  should  be  at  least  1J"  in  thickness,  we  will  make 
each  of  these  bearings  1J"  thick,  although  we  obtain  an  excess  of  bearing. 
However,  as  the  shoe  is  subjected  to  both  bending  and  shear  from  the 
direct  load  (see  Art.  141)  and  wind,  the  metal  is  not  so  overly  excessive 
as  at  first  appears. 

Assuming  one-fourth  of  the  maximum  reaction  on  the  shoe,  which  is 
equal  to  512,000 -f  4  =  128,000*,  as  coming  on  each  bearing  of  the  shoe, 
and  taking  moments  about  the  center  of  bearing  of  the  end  post,  we  have 

128,000  x  2f  =  304,000  inch  Ibs. 

for  the  maximum  bending  moment  on  the  5^"  pin.  The  lever  arm  2f" 
can  be  either  computed  or  scaled  from  the  drawing.  For  the  bending 
stress  on  the  pin,  due  to  this  moment,  we  have 


/  =/-/r^/^4  = 18;600  lbs-  Per  S<1- in-  (about) 


which  shows  the  pin  to  be  amply  strong,  as  25,000*  per  square  inch  is 
allowed.  It  requires  a  moment  of  408,350''*  to  stress  the  5J"  pin  25,000* 
per  square  inch.  The  maximum  bending  moment  allowed  on  the  different 
sizes  of  pins  is  to  be  found  in  tables  of  practically  all  structural  hand- 
books, and,  in  designing  pins,  usually  these  tables  are  consulted  instead 
of  computing  the  fiber  stress  as  is  done  above. 

For  the  maximum  shearing  stress  per  square  inch  on  the  pin  we  have 

128,000  ±  Trr2  =  128,000  -=-  23.75  =  5,390  Ibs.  per  sq.  in., 

which  is  quite  low,  as  12,000*  is  allowed.  As  seen  from  above,  the  pin  is 
really  larger  than  required  and,  theoretically,  should  be  made  smaller; 
but  if  reduced  in  size  the  bearing  would  increase,  little  would  be  saved, 
and,  judging  from  general  appearance,  a  5J"  pin  is  as  small  as  should 
be  used  in  this  class  of  work;  hence  the  assumed  5J"  pin  will  be  used. 

Six-inch  rollers  are  the  minimum  size  permitted  by  the  specifications, 
and  as  that  size  appears  to  be  about  the  correct  size  for  this  case,  they 
will  be  used.  Then,  for  the  linear  inches  of  roller  required,  we  have 

512,000 


600x6 


=  142.2  ins. 


Now,  making  each  roller  30"  long  and  deducting  for  the  7 — £"  slots  in 
pedestals,  we  have  (30  -  7  X  f )  x  6  =  148.5",  which  is  a  few  inches  more 
than  is  required,  but  is  about  as  close  as  we  can  come  and  at  the  same  time 
obtain  good  details  all  around.  So  6 — 6"  rollers,  each  30"  long,  as  shown, 
will  be  used.  The  rollers  are  notched  J"  into  the  cast  shoe  and  pedestal, 
so  as  to  prevent  their  sliding  transversely  (due  to  wind),  and  the  side 
bars  on  their  two  ends  control  their  relative  longitudinal  motion.  It  is 
necessary  that  the  pedestal  and  the  base  of  the  shoe  project  to  clear  the 
side  bars  and  bolt  heads.  In  some  cases  it  is  necessary  for  the  pedestal 
and  the  base  to  project  out  farther  to  obtain  sufficient  bearing  on  the 
masonry.  In  this  case  we  have 


380  STRUCTURAL  ENGINEERING 

for  the  required  area  of  bearing  and  we  have,  as  shown,  (3'  —  4J")  x 
(V  -  4J")  =  l,154n",  which  is  about  300n"  more  than  required,  but  the 
roller  nest  requires  about  the  same  size  of  pedestal  as  shown  and  there 
would  be  but  little  saving,  if  any,  in  reducing  it  to  exactly  the  theoretical 
size.  So  the  size  shown  will  be  used. 

In  drawing  the  lateral  connection  at  LO,  we  first  determine  the  num- 
ber of  J"  field  rivets  necessary  to  develop  the  lateral  in  tension.  The 
lateral  is  a  5"  x  3^"  x  f"  angle,  which  has  a  net  area  of  5.06n"  (allowing 
for  1 — V  rivet  hole).  Multiplying  this  area  by  16,000  and  dividing  by 
6,000  we  obtain  13.5,  say  14  rivets.  As  is  seen,  14  are  used.  The  bolts 
connecting  the  ^§"  lateral  plate  to  the  shoe  should  be  sufficient  to  carry 
the  component  (along  the  end  floor  beam)  of  the  stress  in  the  lateral. 
This  component  is  equal  to  81,000  -f  secco  =  81,000  -r  1.85  =  43,800*  (about) 
(see  Art.  177).  Dividing  this  by  6,000  we  obtain  7.3  for  the  number  of 
J"  turned  bolts  required.  As  is  seen,  7  are  used.  Part  of  the  component 
(43,800*)  may  be  transferred  along  the  floor  beam  to  the  other  shoe,  in 
which  case  less  than  7  bolts  would  be  required ;  but  by  using  the  7  we  are 
sure  that  the  detail  is  sufficiently  strong. 

For  the  component  of  the  stress  in  the  lateral  along  the  bottom  chord 
we  have 

81,000 xsin<o  =  81,000x0.8418  =  68,200  Ibs.  (about). 

Dividing  this  by  6,000  (the  allowable  shear  on  a  J"  field  rivet)  we  obtain 
11.3,  say  12  rivets.  We  have,  as  shown,  14.  However,  some  of  these 
are  also  needed  to  transmit  the  direct  stress  of  the  bottom  chord  into  the 
gusset  plate,  but  as  the  maximum  stress  in  the  bottom  chord  and  lateral 
are  not  likely  to  occur  at  the  same  time,  the  detail  as  shown  will  be 
considered  sufficient.  The  shop  rivets  connecting  the  2 — 4"  x  4"  angles 
to  the  TV'  lateral  plate  should  be  sufficient  to  transmit  the  68,200*  com- 
ponent in  bearing  on  the  iV'  plate. 

One  J"  rivet  at  24,000*  bearing  on  a  TV'  plate  will  transmit 
£  x  T7F  x  24,000  =  9,188*.  Then  we  have  68,200  -f-  9,188  =  7.4  for  the  num- 
ber of  rivets  required,  and,  as  is  seen,  8  are  used.  The  number  of  rivets 
required  to  connect  the  •£%"  lateral  plate  to  the  bottom  of  the  end  floor 
beam  should  be  sufficient  to  take  at  least  one-half  of  the  component  along 
the  floor  beam  of  the  stress  in  the  lateral.  This  requires  about  4  rivets, 
but  to  obtain  a  good,  rigid  connection,  more  are  needed.  So  8  are  used, 
wherein  we  are  governed  very  much  by  appearance  and  sense  of  fitness. 

The  3J"  longitudinal  holes  in  the  cast-steel  pedestal  should  be  cored 
in  casting  and  the  J"  slots  cut  at  the  machine  shop.  If  the  slots  were  to 
be  made  in  casting,  the  pedestal  would  be  likely  to  warp  in  cooling.  These 
slots  and  holes  in  the  pedestal  are  for  two  purposes :  one  is  to  save  metal ; 
and  the  other  one  is  to  permit  dust  and  dirt  that  would  accumulate  between 
the  rollers  to  fall  down  into  the  holes.  In  other  words,  this  arrangement 
permits  the  dirt  that  is  blown  into  the  roller  nest  to  be  blown  out,  and  thus 
the  roller  nest  is  not  likely  to  become  clogged  with  dirt  and  rust.  The 
side  bars  at  each  end  of  the  rollers  should  be  at  least  ff"  thick  to  be  of 
much  service. 

Joint  LI.  Taking  the  next  lower  chord  joint  LI  (Fig.  283),  the 
cross-section  of  the  floor  beam  and  end  connection  angles  of  the  same  can 
be  drawn  as  shown,  and  the  outline  of  the  bottom  chord  collision  strut  and 


381 


382  STRUCTURAL  ENGINEERING 

the  hanger  can  be  sketched — care  being  taken  to  provide  about  \"  clear- 
ance between  these  members.  Then  the  rivet  lines  can  be  drawn  and  the 
rivets  spaced  as  shown.  The  stress  in  the  collision  strut  as  given  in  Art. 
181  is  93,000*.  Dividing  this  by  6,000*  we  obtain  15.5,  say,  16 — £•" 
field  rivets  or  bolts,  as  bolts  should  be  used  in  this  particular  connection. 
As  is  seen,  16  are  used.  Dividing  the  maximum  end  shear  on  the  floor 
beam,  which  is  given  on  the  stress  sheet  (Fig.  279)  as  187,300,  by  6,000 
we  obtain  about  31  rivets  for  the  number  of  J"  field  rivets  required  to  con- 
nect the  floor  beam  to  the  truss.  All  of  these  should  be  above  the  bottom 
chord,  as  there  is  no  diaphragm  in  the  bottom  chord  to  transmit  the 
pressure  across  to  the  other  side  of  the  hanger  and,  consequently,  the 
rivets  shown  connecting  the  floor  beam  to  the  bottom  chord  must  not  be 
considered  to  carry  any  of  the  end  shear  on  the  floor  beam.  As  is  seen, 
there  are  32  rivets  connecting  the  floor  beam  directly  to  the  hanger, 
which  is  really  one  more  than  required  theoretically. 

After  spacing  the  rivets  in  the  collision  strut  and  floor  beam,  the 
gusset  plate  can  be  drawn  to  suit  this  spacing  as  shown.  The  rivets 
connecting  the  gusset  plate  to  the  bottom  chord  should  at  least  be  suffi- 
cient to  transmit  the  component  of  the  stress  in  the  collision  strut  which 
is  equal  to  93,000  x  sin0  =  59,500*.  Dividing  this  by  7,200  we  obtain  a 
little  over  8 — J"  shop  rivets,  or  4  on  a  side,  whereas  12  (not  including 
the  field  rivets  passing  through  the  floor  beam)  are  used,  but  the  detail 
is  satisfactory,  as  this  number  is  necessary  to  obtain  a  well-balanced 
joint.  Next,  the  bottom  lateral  connections  can  be  drawn.  The  number  of 
rivets  required  in  the  lateral  to  the  left  of  LI  is  14 — J"  field  rivets,  as 
determined  at  LO,  and  the  detail  can  be  drawn  as  shown.  The  lateral  on 
the  right-hand  side  of  LI  (Fig.  283)  is  made  of  1— 5"  x  3J"  x  J"  angle, 
which  will  have  a  net  area  of  3.5n".  Then  for  the  strength  of  this  lat- 
eral we  have  3.5x16,000  =  56,000*.  Dividing  this  by  6,000  we  obtain 
about  9 — J"  field  rivets.  This,  as  is  seen,  is  the  number  used.  The  num- 
ber of  rivets  connecting  the  lateral  plate  to  the  floor  beam  should  be  suffi- 
cient to  transmit  the  component  of  the  lateral  to  the  left  of  LI  along  the 
floor  beam.  This  component  is  equal  to  5.06  x  16,000  x  cosw  =  80,960  x 
0.5397  =  43,700*.  Dividing  this  by  6,000  we  obtain  about  7— I"  field 
rivets,  whereas  10  are  used.  After  the  rivets  are  spaced  in  the  laterals 
the  lateral  plate  can  be  drawn  to  suit;  that  is,  so  the  edges  at  no  point 
will  be  less  than  1-J"  from  the  rivets.  Then  the  rivets  connecting  the  lat- 
eral plate  to  the  floor  beam  can  be  spaced  as  shown. 

Joint  L2.  We  will  next  consider  lower  chord  joint  L2  (Fig.  283). 
The  floor  beam  connection  at  this  joint  must  be  exactly  the  same  as  at 
LI,  as  the  floor  beams  should  be  the  same  throughout.  However,  if  the 
spacing  established  at  LI  does  not  suit  at  L2,  it  must  be  changed  so  it 
will  be  satisfactory  for  the  two  joints  and,  likewise,  for  the  other  joints. 
In  each  case  there  must  be  32  rivets  above  the  chord  as  was  determined 
at  LI.  After  drawing  the  cross-section  of  the  floor  beam  and  the  end 
connection,  as  shown  (Fig.  283),  the  outlines  of  the  bottom  chord  and 
diagonal  can  be  drawn.  The  net  area  of  the  main  diagonal,  as  given  on 
the  stress  sheet  (Fig.  279),  is  23.9n".  Multiplying  this  by  16,000  and 
dividing  by  6,000  we  obtain  about  64 — J"  field  rivets  for  the  end  con- 
nection, or  32  on  a  side,  which,  as  is  seen,  is  the  number  used.  There 
should  be  a  sufficient  number  of  rivets  connecting  the  gusset  plates  to 


DESIGN  OF  SIMPLE  KAILKOAD  BEIDGES  383 

the  post  to  transmit  the  vertical  component  of  the  stress  in  the  diagonal, 
which  is  equal  to  23.9  x  16,000  -=-sec0  =  294,000*.  Dividing  this  by  6,000 
we  obtain  49 — J"  field  rivets  or,  say,  24  on  a  side.  As  is  seen,  24  on  a 
side  are  used. 

There  should  be  a  sufficient  number  of  rivets  to  the  left  of  the  floor 
beam  connecting  the  gusset  plate  to  the  chord  (at  L2)  to  transmit  the 
horizontal  component  of  the  stress  in  the  diagonal.  For  this  component 
we  have  23.9  x  16,000  x  sin(9  =  244,700*,  and  dividing  by  7,200  we  obtain 
34 — y  rivets  (shop),  or  17  rivets  on  each  side  of  the  chord,  whereas  16 
are  used,  but  the  rivets  passing  through  the  floor  beam  and  those  to  the 
right  of  the  floor  beam  can  be  counted  on  for  more  than  making  up  for  the 
one  rivet  missing. 

The  splice  in  the  bottom  chord  just  to  the  left  of  L2  should  be  suffi- 
cient to  develop  the  strength  of  the  bottom  chord  member  Z/l-Z/2.  The 
chord  is  spliced,  as  is  seen,  by  the  12"  x  TV  inside  plates,  by  the  12"  X  J" 
outside  splice  plates  and  the  f "  tie  plates.  The  rivets  through  the  web 
of  the  channels  to  the  left  of  the  splice,  which  are  field  rivets,  are  in  double 
shear  and  bearing  on  the  web  of  the  channels.  The  strength  of  the  chan- 
nels is  equal  to  20.9  x  16,000  =  334,000*,  or  167,000*  for  each  channel. 
Then  considering  one  channel,  the  16  field  rivets  in  the  web  are  good  for 
192,000*  (=2x6,000x16)  in  double  shear,  and  145,600*  (=0.52  xjx 
16  x  20,000)  in  bearing  on  the  web  of  the  channel;  and  hence  the  latter 
governs.  The  6  rivets  in  the  flanges  of  the  channel  are  good  for 
6x6,000  =  36,000*.  Adding  this  to  the  145,600*  we  have  181,600*, 
which  is  14,600*  more  than  is  necessary  to  provide  for ;  or,  in  other  words, 
there  is  at  least  one  rivet  too  many;  but  the  rivets  in  the  tie  plates  are  none 
too  numerous  to  hold  those  plates  and  to  omit  one  rivet  in  the  web  would 
leave  the  spacing  unbalanced;  that  is,  unsymmetrical,  and  to  omit  two 
would  be  too  much  of  a  reduction,  so  the  rivets  as  shown  will  be  used. 
There  should  be  enough  rivets  in  the  12"  x  \"  splice  plate  on  each  side 
of  the  splice  to  develop  the  strength  of  that  plate.  The  net  section  of  the 
plate  is  equal  to  12  x  \  -  1  =  5n".  Multiplying  this  by  16,000  we  have 
80,000*  for  the  strength  of  the  plate.  Dividing  this  by  7,200  we  obtain 
11  for  the  number  of  required  shop  rivets,  and  dividing  it  by  6,000  we 
obtain  13.3,  say  14,  for  the  number  of  required  field  rivets.  As  is  seen, 
we  have  12  shop  rivets  on  one  side  of  the  splice  and  16  field  rivets  on 
the  other  side,  so  that  the  riveting  as  far  as  the  outside  splice  plates  are 
concerned  is  quite  satisfactory. 

The  net  area  of  the  12"  x^"  plate  is  5.25- 0.87  =  4.38n",  and  the 
net  area  of  the  12"  xj"  plate  is  5n",  making  a  total  of  4.38  +  5  =  9.38n" 
in  the  splice  plates,  while  the  net  area  of  the  15"x40*  channel  is 
10.45n" ;  but  the  area  of  the  tie  plates  can  be  counted  to  the  extent  of  6 
field  rivets.  This  is  equal  to  6,000  x  6  -r  16,000  =  2.25n".  Adding  this  to 
the  area  of  the  splice  plates  we  obtain  11.63n"  for  the  total  net  area  of 
the  plates  splicing  each  channel — which  is  quite  sufficient. 

The  number  of  rivets  in  the  laterals,  shown  at  L2  (Fig.  283),  is 
obtained  by  developing  the  laterals.  For  example,  the  lateral  to  the  left 
of  the  floor  beam  is  a  5"  x  3J"  x  \"  angle  which  has  a  net  section  of 
4-0.5  =  3.5n".  Multiplying  this  by  16,000  we  obtain  56,000*  for  the 
strength  of  the  lateral,  and  dividing  this  by  6,000  we  obtain  9 — J"  field 


384 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES  385 

rivets,  whereas  9  are  used.  The  number  of  rivets  in  the  other  laterals  is 
obtained  in  the  same  manner. 

Joint  L3.  Next  we  will  make  a  sketch  of  the  details  at  the  joint 
L3,  as  shown  in  Fig.  284.  In  drawing  this  sketch,  we  can  draw  the 
outline  of  the  bottom  chord,  diagonals  and  post.  Then  we  can  draw  the 
end  connections  of  the  floor  beam  onto  the  sketch  and  draw  the  rivet 
lines  in  all  the  members,  and  then  we  are  ready  to  determine  the  number 
of  rivets  and  spacing  of  same  for  each  member  connecting  at  that  joint. 

The  rivets  in  the  connection  of  the  two  diagonals  will  be  the  same 
for  each.  These  diagonals  are  subjected  to  both  tension  and  compres- 
sion, and,  according  to  the  specifications,  the  sum  of  the  two  stresses 
should  be  used  in  determining  the  required  number  of  rivets  in  the  end 
connections.  So  we  have  (201,000  + 78,000)  =279,000*  for  the  stress  to 
be  considered  (see  Fig.  279).  Dividing  this  by  6,000  we  obtain  46.5— J" 
field  rivets  or,  say,  24  on  each  side  of  each  diagonal,  arid,  as  is  seen,  24 
are  used.  The  rivets  on  each  side  of  the  floor  beam  connecting  the 
gusset  plate  to  the  bottom  chord  should  be  sufficient  to  transmit  the  com- 
ponent of  the  stress  in  the  diagonal  along  the  bottom  chord.  This  com- 
ponent is  equal  to  279,000*  x  sin  0  =  279,000x0.64  =  178,000*  (about). 
Dividing  this  by  7,200*  (the  allowable  single  shear  on  a  J"  shop  rivet) 
we  obtain  24.8  rivets  or,  say,  13  on  a  side  and,  as  is  seen,  14  are  used. 
The  rivets  in  the  diagonals  should  be  spaced  first  and  the  edge  of  the 
gusset  plate  located  and  drawn  down  to  the  bottom  chord,  and  the  rivets 
arranged  in  the  bottom  chord  to  suit  the  gusset  plate.  Then  the  next 
thing  to  do  is  to  determine  the  rivets  connecting  the  floor  beam  to  the 
truss,  as  was  explained  above  in  the  case  of  panel  points  IA  and  L2. 

After  having  the  rivets  spaced  in  the  floor  beam  connection,  the 
gusset  plate  can  be  drawn  completely;  however,  the  spacing  in  the  floor 
beam  connection  selected  for  this  joint  must  be  the  same  as  for  the  other 
intermediate  lower  chord  joints.  The  bottom  lateral  connections  can 
next  be  drawn.  The  rivets  in  each  lateral  are  determined  by  developing 
the  strength  of  each  lateral  as  was  shown  above  for  the  other  joints. 
After  the  rivets  are  spaced  in  the  laterals  the  lateral  plate  can  be  drawn 
to  suit  the  spacing;  however,  the  rivets  in  the  bottom  of  the  floor  beam 
should  be  the  same  as  required  at  LI,  in  order  to  have  all  of  the  inter- 
mediate floor  beams  alike. 

Referring  to  the  sketch  of  the  intermediate  floor  beam,  drawn  just 
to  the  right  of  L3  (Fig.  284),  the  rivets  connecting  the  end  connection 
angles  to  the  floor  beam  must  be  sufficient  to  transmit  the  maximum  end 
shear  on  the  floor  beam  in  double  shear  or  bearing,  whichever  requires 
the  greater  number.  These  end  angles  are  placed  upon  \"  fillers  (the 
fillers  being  just  as  thick  as  the  flange  angles).  These  fillers  are  held 
by  rivets  placed  outside  of  the  connection  angles,  and  hence  the  thickness 
of  bearing  on  the  rivets  through  the  angles  can  be  taken  as  the  total 
thickness  of  the  2 — |"  fillers  and  web,  making  lyV'j  or  *ne  combined 
thickness  of  the  two  angles,  which  is  the  least  thickness — as  it  is  1".  So 
for  the  allowable  bearing  on  each  J"  shop  rivet  in  this  connection  we 
have  Jx  1x24,000  =  2 1,000*,  and  for  the  allowable  double  shear  on  the 
same  we  have  2x0.6x12,000  =  14,400*,  which  is  less  than  the  bearing 
and  hence  governs.  Then  dividing  the  maximum  end  shear  by  the  last 


386  STRUCTURAL  ENGINEERING 

figure  we  have  187,300  +  14,400  =  13  rivets,  and,  as  is  seen,  13  rivets  are 
used,  not  counting  those  passing  through  the  flange  angles. 

There  should  be  enough  rivets  connecting  the  stringer  to  the  floor 
beam  to  transmit  the  maximum  end  shear  on  a  stringer.  So  we  have 
142,000  +  6,000  =  23.6  rivets  in  single  shear,  and,  as  is  seen,  24  are  used. 
The  bearing  of  these  rivets  is  upon  the  2 — \"  fillers  and  the  TV  web,  so 
that  the  shear  governs.  There  should  be  enough  rivets  through  these 
fillers,  outside  of  the  stringer  connection,  to  carry  the  total  end  shear  on 
a  stringer  and  also  the  concentration  on  the  floor  beam.  In  the  first 
case,  the  rivets  are  in  single  shear  and  for  the  number  required  we  have 
142,000  4- 7,200  =  about  20  shop  rivets,  whereas  we  have  20.  In  the  sec- 
ond case,  these  rivets  are  in  bearing  on  the  -£%"  web  and  for  the  number 
required  we  have  187,300  -=-9,180  =  about  20  rivets,  so  the  correct  number 
is  used.  As  a  matter  of  fact,  the  field  rivets  in  the  stringer  connection 
(theoretically)  bear  on  the  web  also  and  hence  we  have  an  excess  of  24 
field  rivets  in  bearing  on  the  web  in  the  connection  shown,  but  it  is  a 
question  in  the  case  of  such  long  rivets,  and  especially  when  field  driven, 
just  how  much  bearing  they  will  exert  on  the  web,  and  there  is  some 
question  as  to  the  bending,  and  to  be  on  the  safe  side  it  is  advisable  to 
compute  the  number  on  this  important  connection  as  is  here  shown. 

It  will  be  seen  that  in  the  case  of  the  end  connection  of  the  floor 
beam  there  is  an  excess  of  5  rivets  in  bearing  on  the  web  (not  considering 
the  ones  passing  through  the  flange — they  resist  the  flange  increment) 
and  consequently  5  rivets  could  (theoretically)  be  omitted  from  the  fillers, 
but  this  would  make  the  rivets  look  rather  sparing,  so  the  number  shown 
will  be  used. 

The  rivets  through  the  fillers  of  the  stringer  connection  are  counter- 
sunk on  one  side  of  the  connection  to  permit  the  swinging  of  the  stringers 
in  position  during  the  erecting  of  the  structure.  The  small  angles  under 
each  stringer  (connected  to  the  floor  beam)  are  for  erection  purposes 
and  are  not  assumed  to  carry  any  load  from  the  stringers. 

Joint  £71.  We  will  next  consider  the  drawing  of  the  hip  joint  Ul 
shown  in  Fig.  285.  The  first  thing  to  do  in  making  this  sketch  is  to 
select  the  center  point  of  the  joint  and  draw  the  center  lines  of  the  end 
post  and  top  chord.  Then  bisect  the  angle  between  these  lines,  and  this 
bisector  will  be  the  joint  line — where  the  two  members  meet.  Now,  using 
the  same  transverse  spacing  as  was  used  in  the  end  post  at  LO  (Fig. 
282),  the  outlines  of  the  end  post  and  top  chord  can  be  drawn  and  the 
rivets  spaced.  There  should  be  enough  rivets  connecting  the  end  post 
to  the  gusset  plates  to  transmit  the  total  stress  in  the  member.  So 
we  have 

523,000  +  6,000  =  87,  say  88,— J"  field  rivets, 

or  44  on  a  side.  As  is  seen,  there  are  38  in  single  shear,  which  provides 
for  6,000x38  =  228,000*  of  the  stress,  and  4  passing  through  the  small 
outside  splice  plate  which  can  be  considered  for  bearing  on  the  \"  web 
and  hence  provide  for  8,750  x  4  =  35,000*  of  the  stress  in  the  end  post. 
Adding  these  two  stresses  we  have  228,000  +  35,000  =  263,000*,  and  mul- 
tiplying this  by  2  we  have  526,000,  which  is  3,000*  more  than  the  stress 
in  the  end  post,  and  hence  the  riveting  as  shown  in  the  end  post  is  suffi- 
cient. There  should  be  enough  rivets  connecting  the  top  chord  to  the 


387 


STRUCTURAL  ENGINEERING 


gusset  plates  to  transmit  the  total  stress  in  that  member.  As  is  shown, 
there  are  33  shop  rivets  (J"  diameter)  which  provide  for  33x7,200  = 
237,600*  of  the  stress  in  the  top  chord,  and  the  4  field  rivets  in  bearing 
on  the  f"  web  which  provide  for  6,563  x  4  =  26,250*  stress.  Adding  these 
two  quantities  together  and  multiplying  by  2  we  have  263,850x2  = 
527,700*,  which  shows  this  riveting  to  be  satisfactory,  as  the  stress  in 
the  top  chord  UI-L2  is  521,000*  (see  Fig.  279).  Next,  the  hanger 
Ul-Ll  and  the  diagonal  U1-L2  can  be  drawn  and  the  rivets  connecting 
them  to  the  gusset  plates  spaced  as  shown.  The  rivets  connecting  the 
hanger  to  the  gusset  plates  should  be  sufficient  to  develop  the  strength 
of  the  hanger,  which  is  12.94x16,000  =  207,000*  (see  Fig.  279).  Then 
we  have 

207,000-6,000  =  34.5,  say  36,— J"  field  rivets, 

or  18  on  a  side.  As  is  seen,  18  are  used.  The  number  of  rivets  con- 
necting the  diagonal  U1-L2  to  the  gusset  plates  should  be  sufficient  to 
develop  the  strength  of  the  diagonal,  which  is  23.9x16,000  =  382,000*. 
Dividing  this  by  6,000*  we  obtain  64 — £"  field  rivets,  or  32  on  a  side; 
whereas  32  are  used,  as  is  seen. 

There   should   be   enough   metal   in   the   gusset   plates   to   properly 
transmit  all  of  the  forces  acting  in  the  plane  of  the  truss  at  the  joint. 


Fig.  286 

These  forces  are  as  indicated  in  Fig.  286.  As  is  readily  seen,  these 
forces  will  not  all  be  a  maximum  at  the  same  time.  About  the  most 
severe  stress  will  occur  on  the  plates  when  the  end  post  has  a  maximum 
stress,  as  the  top  chord  U1-U2  has  about  the  maximum  stress  at  that  time 
and  the  hanger  Ul-Ll  has  quite  a  large  stress,  and  the  diagonal  U1-L2 
has  a  low  stress.  Combining  S  and  S3  (Fig.  286)  we  obtain  the  resultant 
R2,  and  combining  SI  and  S2  we  obtain  the  resultant  Rl,  which  is  equal 
and  opposite  to  R2.  So,  evidently,  the  maximum  tension  on  the  plates 
will  be  on  same  section  as  cd,  and  combining  S  and  S2  we  obtain  the 
resultant  R3 ;  and,  likewise,  combining  SI  and  S3  we  obtain  the  resultant 
7?4,  which  will  be  equal  and  opposite  to  R3.  So,  evidently,  the  maximum 


DESIGN  OF  SIMPLE  EAILKOAD  BEIDGES  389 

compression  on  the  gusset  plates  will  be  on  some  section  as  ab  perpen- 
dicular to  these  resultants  (R3  and  .R4),  due  mostly  to  S  and  £3.  A 
very  satisfactory  approximation  can  be  obtained  by  making  the  plates 
thick  enough  so  that  two-thirds  of  the  metal  along  the  section  ab  is  suffi- 
cient to  transmit  the  maximum  stress  in  the  end  post.  By  making  the 
plate  f "  thick,  two-thirds  of  the  section  along  section  ab  is  about  48  x  f  x 
f  x2  =  40n".  Dividing  this  into  the  maximum  stress  in  the  end  post,  we 
have  523,000  -h40n"  =  13,100#  (about)  for  the  maximum  compressive 
unit  stress  in  the  plates,  which  is  about  the  correct  value,  as  it  is  about 
the  same  as  allowed  in  the  chord  U1-U2.  So  the  gusset  plates  will  be 
made  f"  thick.  The  same  thickness  will  be  used  at  all  of  the  other 
joints,  whether  required  or  not,  in  order  to  have  the  sides  of  the  truss 
members  in  the  same  plane  throughout  without  the  use  of  fillers. 

After  the  connections  of  the  main  members  at  joint  £71  are  drawn 
as  shown  (Fig.  285),  the  portal  can  be  drawn.  In  drawing  the  portal, 
the  first  thing  to  do  is  to  draw  a  top  plan  of  the  end  post  and  locate 
the  center  of  the  portal.  Then  locate  the  clearance  line  as  shown  and 
draw  the  cross-section  of  the  portal  on  the  elevation  of  the  end  post  (as 
shown),  and  then  draw  the  plan  of  the  portal,  keeping  inside  the  clear- 
ance line  at  every  point. 

The  required  number  of  rivets  in  the  end  connections  of  the  portal 
members  is  obtained  by  developing  the  strength  of  each  of  the  members. 
After  the  portal  is  drawn  the  top  lateral  and  the  bent  lateral  plates  can 
be  drawn  as  shown,  and  thus  the  sketch  of  joint  £71  is  completed. 

Joint  U2.  The  details  of  this  joint  are  shown  in  Fig.  287.  The 
outlines  of  the  top  chord  as  seen  in  elevation  should  be  drawn  first,  and 
then  the  outlines  of  the  diagonal  (C72-L3)  and  the  post  (C72-Z/2)  can  be 
drawn  as  shown.  The  next  thing  after  this  is  to  locate  the  rivets  in  the 
end  connection  of  the  diagonal.  As  this  diagonal  is  subjected  to  both 
tension  and  compression,  the  number  of  rivets  in  each  end  connection 
must,  according  to  the  specifications,  be  sufficient  to  transmit  the  sum  of 
the  two  stresses.  So  we  have 

(78,000  +  201,000)  -r  6,000  =  46.5— I"  rivets 

for  the  required  number,  or  say  24  on  a  side,  and,  as  is  seen,  24  are 
used,  the  same  as  in  the  end  connection  of  this  same  diagonal  at  L3  (see 
Fig.  284). 

Next,  the  cross-section  of  the  transverse  strut  can  be  drawn  on  to 
the  chord  and  then  the  transverse  view,  shown  to  the  right,  can  be 
drawn.  In  drawing  this  view  it  is  best  to  draw  first  the  cross-section 
of  the  top  chord,  as  shown,  and  then  the  elevation  of  the  transverse  strut. 
Next  the  clearance  line  should  be  located  and  dotted  in,  as  shown,  and 
then  the  knee  brace  and  sub-strut  can  be  drawn,  at  which  time  the  con- 
nections of  the  knee  brace  and  sub-strut  to  the  post  are  determined  and 
can  be  projected  over  to  the  other  view  of  the  post  as  shown.  The  next 
thing  to  do  is  to  determine  the  number  of  rivets  required  to  connect  the 
post  (C72-L2)  to  the  gusset  plates. 

This  member  is  subjected  to  162,000*  compression  and  54,000* 
tension.  So,  according  to  the  specifications,  there  should  be  a  sufficient 
number  of  rivets  connecting  the  post  to  the  gusset  plates  at  U2  to  trans- 
mit the  sum  of  these  two  stresses.  So  we  have  (162,000  +  54,000) -5- 


J 


Fig.  287 


390 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES  391 

6,000  =  36— J"  field  rivets,  or  18  on  each  side  of  the  post,  whereas  20 
are  used. 

There  should  be  enough  rivets  (at  least)  connecting  the  gusset 
plates  to  the  top  chord  to  transmit  the  horizontal  component  of  diagonal 
U2-L3.  This  component,  considering  the  sum  of  the  stresses  in  the  diag- 
onal, is  equal  to  (201,000  +  78,000)  x  sin  (9  =  279,000  x  0.64  =  178,560*. 
Dividing  this  by  7,200  we  obtain  about  25  shop  rivets,  say  13  on  a  side, 
whereas  there  are  almost  three  times  this  number  shown  (Fig.  287). 
But  the  number  cannot  well  be  reduced,  as  the  spacing  of  the  rivets  in 
the  chord  angles  is  about  what  the  specifications  require,  and  the  rivets 
between  the  chord  angles  cannot  be  more  sparingly  spaced.  So  the  rivet- 
ing of  the  gusset  plates  to  the  chord  will  be  considered  satisfactory  as 
shown. 

The  next  thing,  the  plan  of  the  top  chord,  strut  and  laterals  can  be 
drawn.  There  should  be  enough  rivets  in  each  lateral  to  develop  the 
strength  of  the  lateral  in  tension.  Each  lateral  is  composed  of  2 — Ls 
3J"  x  3J"  x  f".  So  for  the  strength  of  each  we  have  (4.96- 0.75)  x 
16,000  =  67,360*.  Dividing  this  by  6,000  we  obtain  about  11  field  rivets, 
say  5  in  each  angle,  whereas  5  are  used — 5  in  the  angle  shown  and  5  in 
the  angle  at  the  bottom  of  the  chord. 

There  should  be  enough  rivets  in  each  transverse  strut  to  develop 
the  strength  of  the  strut  in  compression.  Each  of  these  struts  is  com- 
posed of  4 — Ls  3|-"  x  3"  x  f "  arranged  so  as  to  form  an  I-section.  So, 
considering  the  3J"  legs  turned  horizontally  and  the  vertical  legs  -J", 
back  to  back,  we  have 

1 Q9 
p  =  16,000  -  70  j~  =  8,140  Ibs. 

for  the  allowable  unit  stress.  Then  multiplying  this  by  the  area  of  the 
4  angles  in  each  strut  we  obtain  8,140x9.2  =  74,888.  Dividing  this  by 
6,000  we  obtain  about  13  field  rivets,  say  7  in  the  top  angles  and  7  in 
the  bottom  angles,  whereas  8  are  used  so  as  to  have  symmetrical  spacing. 

The  chord  splice  just  to  the  left  of  joint  U2  (Fig.  287)  is  what  is 
known  as  a  butt  joint.  The  chord  sections  are  planed  so  they  fit  per- 
fectly against  each  other  and  consequently  the  stress  can  be  considered 
as  being  transmitted  from  one  chord  segment  to  the  other  without  the  aid 
of  the  splice  plates,  and  hence  the  splice  plates  are  considered  as  merely 
holding  the  chord  segments  in  position ;  and,  as  is  evident,  the  size  of  these 
plates  and  the  number  of  rivets  connecting  them  to  the  chord  is  mostly 
a  matter  of  judgment. 

Joint  U3.  The  details  at  this  joint  will  be  practically  the  same  as 
at  C72,  except  the  connection  of  the  diagonal  and  chord  splice  are 
omitted,  and  consequently  no  larger  scale  sketch  is  necessary. 

This  completes  the  necessary  large  scale  sketches  and  next  the  gen- 
eral drawing,  Fig.  288,  and  the  shop  drawings,  Figs.  289  to  303,  can  be 
made,  wherein  the  details  are,  as  we  may  say,  simply  transferred  from 
the  above  sketches  to  these  drawings. 

185.  Camber. — To  prevent  bridge  trusses  from  sagging  they  are 
"cambered,"  that  is,  they  are  built  so  that  they  curve  upward.  In  the 
case  of  an  ordinary  truss  bridge  the  cambering  of  the  trusses  is  accom- 


Fig.  288 


392 


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DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  409 

plished  by  simply  making  the  top  chord  longer  than  the  bottom  chord 
and  increasing  the  length  of  the  diagonals  to  correspond. 

The  top  chords  are  usually  increased  £"  for  each  10  feet  of  length 
(horizontal).  The  length  of  each  diagonal  is  computed  by  taking  the 
mean  of  the  top  and  bottom  chord  lengths  (in  the  corresponding  panel) 
as  the  base  of  a  right-angled  triangle,  and  the  height  of  the  adjoining 
post  as  the  altitude,  and  the  diagonal  as  the  hypotenuse.  This  method 
of  cambering  is  satisfactory  for  trusses  up  to  300  feet  in  length,  and 
hence  is  used  in  the  case  of  the  above  bridge.  The  panel  length  here  is 
25'-0",  so  by  increasing  the  top  chord  J"  for  each  10  feet  of  length  we 
obtain  S5'-0-^"  for  the  cambered  length  of  the  top  chord  in  each  panel 
as  is  shown  for  the  chord  sections  in  Figs.  292  and  294.  The  end  posts 
are  not  increased  in  length.  For  the  mean  lengths  of  the  top  and  bottom 
chords  in  each  panel  we  have 


Then  for  the  cambered  length  of  each  diagonal  we  have 


V  (30) a  +  (25'  -  O^y) 2  =  39'  -  °lf  "• 


This  length  is  used  for  the  diagonals,  as  is  seen  in  Fig.  297. 

The  camber  affects  the  lengths  of  only  the  top  chords,  diagonals, 
and  top  laterals. 

186.  Hints  Regarding  Shop  Drawings.  —  Drawing  El  (Fig.  289) 
is  the  erecting  diagram.  This  drawing  is  intended  principally  for  the 
use  of  the  party  erecting  the  structure.  The  mark  (as  is  seen)  of  each 
piece,  or  member,  and  the  general  dimensions  of  the  structure  are  shown 
and  in  addition  a  list  of  the  shop  drawings  is  given. 

Drawing  1  (Fig.  290)  shows  the  complete  details  of  the  stringers 
and  stringer  bracing.  The  work  of  detailing  in  this  case  is  practically 
the  same  as  for  deck  plate  girders  (Art.  136).  There  should  be  enough 
rivets  in  the  flanges  to  transmit  the  flange  increment  and  at  the  same 
time  support  the  vertical  load  transmitted  to  them  from  the  ties.  To 
obtain  the  spacing  of  the  flange  rivets  at  the  ends  of  the  stringers  we  have 

R  =  7,880#,  which  is  the  allowable  bearing  of  a  J"  rivet  on  the 
t"  web; 


v  =  1,200#  =  +  100%  impact  (see  Art.  136)  ; 

S  =  142,000*  ; 
h  =  41.5. 

Then  substituting  these  values  in  Formula   (4),  Art.   116   (the  formula 
required  by  the  specifications),  we  obtain 

7,880 


p  =IWW  +     -  =2.18  (about),  say  2J,  ins. 


410  STRUCTURAL  ENGINEERING 

for  the  spacing  of  the  flange  rivets  near  the  ends  of  the  stringers,  and, 
as  is  seen,  this  spacing  is  used.  The  spacing  at  intermediate  points  is 
obtained  as  explained  in  Art.  136. 

The  rivets  connecting  each  pair  of  end  stiffeners  to  the  stringer 
should  be  sufficient  to  transmit  the  maximum  end  shear  on  the  stringer. 
The  two  angles  have  a  combined  thickness  of  1"  in  bearing  on  the  rivets, 
and  the  two  fillers  and  web  combined  have  a  thickness  of  If"  in  bearing. 
Then,  using  J"  rivets,  we  have 

24,000x1  xf  =  21,000  Ibs. 

for  the  allowable  bearing  on  each  rivet.  The  rivets  are  in  double  shear, 
so  for  the  allowable  shear  on  each  rivet  we  have  12,000  x  0.6  x  2  =  14,400. 
As  is  seen,  double  shear  governs.  Then  for  the  number  of  rivets  required 
in  the  angles  we  have  142,000 -~  14,400  =  9.8  (about),  whereas  11  are  used, 
but  the  rivets  in  the  flanges  should  not  be  counted  for  very  much,  as  they 
take  the  flange  increment.  So  the  riveting,  as  shown,  is  about  correct. 
There  should  be  a  sufficient  number  of  rivets  connecting  the  \n  fillers  to 
the  web  to  transmit  the  maximum  end  shear  in  bearing  on  the  f "  web. 
Hence,  for  the  number  required  we  have  142,000-^7,880  =  18  (about), 
and,  as  is  seen,  18  are  used. 

The  intermediate  stiffeners  are  spaced  in  accordance  with  Art.  118 
(see  Art.  170).  The  spacing  of  the  rivets  in  the  intermediate  stiffeners, 
as  previously  stated,  is  mostly  a  matter  of  judgment  (see  Art.  136). 

There  should  be  a  sufficient  number  of  rivets  in  the  end  of  each 
lateral  to  develop  the  strength  of  the  lateral  in  compression.  The  length 
of  each  lateral  is  about  98".  Then  for  the  allowable  unit  stress  on  each 
lateral  we  have 

no 

p  =  16,000  -  70  ^  =  8,400  Ibs.  (about), 
y  \) 

Then  for  the  strength  of  each  we  have  8,400x2.48  =  20,800*.  Dividing 
this  by  6,000  we  obtain  a  little  over  3 — J"  field  rivets,  and,  as  is  seen, 
3  are  used.  There  should  be  enough  rivets  connecting  each  lateral  plate 
to  the  stringer  to  transmit  the  sum  of  the  components  of  the  two  laterals 
along  the  stringers,  as  the  two  act  in  the  same  direction,  one  being  in 
compression  and  the  other  in  tension.  Each  field  rivet  passing  through 
the  flange  will  take  the  component  exerted  on  it  directly  and  hence  it  is 
a  matter  of  taking  care  of  the  component  of  the  rivets  in  the  lateral 
plate  outside  of  the  flange  angle.  Two  of  these  are  good  for  12,000#. 
Resolving  this  along  the  stringer  we  have  12,000x0.69  =  8,280*.  This 
requires  about  1  shop  rivet,  and,  as  is  seen,  1  shop  rivet  is  used. 

The  connection  of  the  bottom  laterals  and  struts  to  the  bottom 
flanges  of  the  stringers  is  mostly  a  matter  of  judgment,  as  the  stresses 
due  to  traction  are  quite  low  (see  Art.  177).  However,  there  should  be 
enough  rivets  in  these  connections  to  insure  rigidity. 

Drawing  2  (Fig.  291)  shows  the  complete  details  of  the  floor  beams. 
The  end  connections  and  the  stringer  connections,  in  the  case  of  the 
intermediate  floor  beams,  were  previously  designed  (see  Art.  184)  and 
will  not  be  considered  here.  The  number  of  flange  rivets  in  the  inter- 
mediate floor  beams  should  be  sufficient  to  transmit  the  flange  increment. 
To  obtain  the  spacing  of  these  rivets  between  the  truss  and  the  stringer 


DESIGN  OF  SIMPLE  KAILKOAD  BEIDGES  411 

we  have 

r  =  9,190#,  which  is  the  allowable  bearing  of  a  J"  rivet  on  the 


S  =  187,300#; 
h  =  49.5, 

Substituting  these  values  in  Formula  (2)   (Art.  116)  we  obtain 

9,190x49.5      0.          .  ,       . 
P=      !87,3QQ     =  3-4  "»•(<*<«*) 

for  the  required  spacing,  whereas  2J"  spacing  is  used. 

Apparently  this  spacing  could  be  a  little  more  than  2J",  but  as  the 
flange  stress  is  zero  at  the  end  of  the  beam  and  practically  a  maximum 
at  the  stringer  connection,  it  is  obvious  that  there  should  be  a  sufficient 
number  of  rivets  in  each  flange,  between  the  end  of  the  beam  and  the 
stringer  connection,  to  transmit  the  flange  stress.  Then  dividing  the 
flange  stress  (see  stress  sheet,  Fig.  279)  by  9,190  we  have  201,000  -r 
9,190  =  22  (about)  for  the  number  of  rivets  required,  whereas  20  are 
used.  From  this  it  is  seen  that  the  number  shown  is  about  the  correct 
number.  In  all  such  cases,  where  the  distance  is  short,  the  rivet  spacing 
should  be  verified  in  this  manner.  In  most  girders,  however,  the  spacing 
required  by  the  flange  increment  governs. 

The  spacing  of  the  flange  rivets  between  the  stringer  connections 
can  be  6",  the  maximum  spacing  allowed,  as  the  shear  between  the  string- 
ers is  (theoretically)  zero  except  for  the  small  amount  due  to  the  weight 
of  the  part  of  the  floor  beam  intervening. 

To  obtain  the  spacing  of  the  flange  rivets  in  the  end  floor  beam  to 
transmit  the  flange  increment,  between  the  end  of  the  beam  and  stringer 
connection,  we  have 

r=  7,880*,  which  is  the  allowable  bearing  of  a  J-"  rivet  on  the 

2L»  web; 

S  =  143,200*  (see  stress  sheet,  Fig.  279); 
fc  =  49.5. 

Then  substituting  these  values  in  Formula  (2)  (Art.  116)  we  obtain 

7,880x49.5 
*=      143,200     =^ 

for  the  required  spacing.  For  the  number  required  to  transmit  the  flange 
stress  we  have  153,000  -f  7,880  =  20  (about),  and,  as  is  seen,  20  rivets  are 
used  in  the  top  flange  and  hence  the  spacing  shown  is  correct,  although 
it  is  less  than  required  by  the  flange  increment. 

In  the  bottom  flange  there  are  14  rivets  between  the  end  of  the 
beam  and  stringer  connection,  all  of  which  can  be  considered  as  bearing 
on  the  f  "  web,  and  those  passing  through  the  plates  extending  over  the 
flange  angles  can  be  considered  as  being  in  double  shear  in  addition  to 
their  bearing  on  the  f  "  web.  So  we  have 

(14  x  7,880)  +  (4  x  14,400)  =  167,900  Ibs. 

for  the  value  of  the  rivets  transmitting  the  flange  stress,  whereas  this 
stress  is  153,000*,  and  hence  the  rivets  shown  in  the  bottom  flange, 
between  the  end  of  the  beam  and  stringer  connection,  are  quite  satis- 
factory. 


412 


STRUCTURAL  ENGINEERING 


The  spacing  of  the  flange  rivets  between  the  stringer  connections 
can  be  6",  as  the  shear  between  the  stringers  is  practically  zero. 

There  should  be  sufficient  metal  through  the  reduced  portion,  near 
the  end  of  the  end  floor  beam,  to  resist  the  cross  bending  at  that  point. 
Taking  a  section  c-c  (Fig.  282)  we  have  the  metal  shown  in  Fig.  304. 

Taking  moments  about  the  back  of  the  top  flange  angles  we  have 
-      (23.62  x  19.75)  +  (12.37  x  16.62)  +  (8.36  x  1.96)  _ 
X=  44.35 

for  the  distance  from  the  back  of  the  angles  to  the  center  of  gravity  of 
the  section. 

V 

* 


-S'ffs. 


Figr.  304 

Then  for  the  moment  of  inertia  about  the  gravity  axis  x-x  we  have 
the  following: 


8.36x13.56  +30.9    =1,568  for  the  angles; 
12.37  xllV  1,122    =1,13  7  for  the  f"  web; 

23.62  x  4lJ32  +  1,435  =  4411  f  or  the  A"  Plates  ; 
4>ooo 

—  295 
Total  moment  of  inertia  =  —     -  for  rivet  holes. 

4,/&oo 

Then  for  the  maximum  stress  on  the  outer  element  we  have 


This  shows  that  the  metal  at  the  reduced  portion  of  the  end  floor  beams 
is  quite  sufficient,  as  the  allowable  stress  is  16,000#. 

Drawings  3  to  14  (Figs.  292  to  303)  are  self-explanatory.  The 
making  of  these  drawings  is  mostly  a  matter  of  copying  the  details  from 
the  large  scale  sketches  (Figs.  282  to  285  and  287). 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  413 

187.  Summary  of  Weight  of  Metal  and  Cost  of  the  Above 
150-Ft.  Span.— 

12— Stringers    @    4,300*.  .  51,600  Ibs. 

Stringer  bracing  2,120  Ibs. 

2— End  floor  beams @  3,010*.  .  6,020  Ibs. 

5— Intermediate  floor  beams @  3,470*.  .  17,350 Ibs. 

4— End  posts  Ul-LO @  8,850*.  .  35,400  Ibs. 

4— Top  chords  U1-U2 @  4,830*.  .  19,320  Ibs. 

2— Top  chords  U2-U2.  . @  10,890*.  .  61,780  Ibs. 

4— Bottom  chords  LO-L2 @  4,730*.  .  18,920  Ibs. 

2— Bottom  chords  L2-L2 @  10,740*.  .  21,480  Ibs. 

4— Hangers  U1-L1  @  1,780*..  7,120 Ibs. 

6— Int.  posts  U2-L2  and  U3-L3  @  2,840*..  17,040 Ibs. 

4— Diagonals  U1-L2  @  3,760*..  15,040 Ibs. 

4— Diagonals  U2-L3  @  2,880*.  .  11,520  Ibs. 

4— Collision  struts  MI-LI @  950*.  .  3,800  Ibs. 

2— Portal  struts  @  2,270*. .  4,540  Ibs. 

Top   laterals    5,630  Ibs. 

Transverse  struts    2,770  Ibs. 

Bottom  laterals  6,460  Ibs. 

4— Shoes  @  940*.  .  3,760 Ibs. 

2— Roller  nests @  1,180*.  .  2,360  Ibs. 

2 — Cast  pedestals  @  1,130*.  .  2,260  Ibs. 

2— Pedestals  @  1,090*.  .  2,180  Ibs. 

Pins    640  Ibs. 

Anchor  bolts  130  Ibs. 

Total  weight  of  bridge 279,240  Ibs. 

Cost  of  span  @  3Jtf  P^  po™d  «  279,240  x  0.0325  =  $9,075.30.  This 
price  (3J^)  per  pound  is  only  a  fair  average  price  for  this  class  of  work. 
The  price  will  vary  from  2f  $  to  4^  per  pound. 

The  effective  dead  load  from  the  metal  is  equal  to  the  total  weight 
of  metal  in  the  span  minus  the  weight  of  the  end  floor  beams,  shoes, 
roller  nests,  pedestals,  pins,  and  anchor  bolts.  So  we  have  379,240- 
17,350  =  261,890*  for  the  effective  dead  weight  of  metal  in  the  span. 
Dividing  this  by  150,  the  length  of  the  span,  we  have  261,89o -=- 150  = 
1,745*  per  foot  of  span,  which  is  only  35*  more  than  assumed  above  (see 
stress  sheet,  Fig.  279).  So  the  dead  load  assumed  is  quite  satisfactory, 
as  10  per  cent  variation  would  not  materially  affect  the  design. 

DRAWING  ROOM  EXERCISE  NO.  7 

Design  a  single-track  through  riveted  railroad  bridge  and  make  a 
stress  sheet  for  same — the  stress  sheet  to  be  upon  tracing  cloth. 
Data: — 

Length  of  span  =  6  panels  at  26'-0"  =  156'-0". 
Height  of  trusses  =  31'-0"  c.c.  of  chords. 
Live  load,  Cooper's  JE50. 
Dead  load,  to  be  assumed. 
Specifications,  A.  R.  E.  Ass'n. 


5-f?ol/ers4l2-4.'tg.  0 


'/'/^s. 

'7&/>£o//* 

^-oi"  %  Turned. 


.ZTteP/s.  /3"*$"*/-e"  _,•  .<• 

/^^^^^  ^Z%8 

'  ""    "          .   /     \   >'     '.  /      X  ,'      "*'/' 


ft*****. 


"rCastSket  Mas.  Plate. 


\  t    'tt/tSi'** 

X  *0  /*/Jvr*.'f 


Fig.  305 


414 


DESIGN  OF  SIMPLE  BAILKOAD  BEIDGES  415 

DRAWING  ROOM   EXERCISE   NO.   8 

Make  a  general  drawing  of  the  above  156-ft.  bridge  —  the  finished 
drawing  to  be  upon  tracing  cloth  and  similar  to  the  one  shown  in 
Fig.  288. 

188.  Remarks.  —  End  floor  beams  are  sometimes  omitted  in  through 
truss  bridges  in  which  case  the  end  stringer  rests  directly  upon  the  piers 
as  shown  in  Fig.  305.     The  ends  of  the  stringers  are  usually  constructed 
as  shown  at  (a).     The  stringers,  as  is  seen,  are  braced  to  each  other  by 
a  cross-frame  and  they  are  connected  to  the  shoes  and  trusses  by  a  trans- 
verse strut  which  forms  the  lower  part  of  the  cross-frame. 

Circular  rollers  are  sometimes  used.  The  roller  nests  in  that  case 
are  usually  constructed  as  shown  in  Fig.  305.  The  A.  R.  E.  Association 
Specifications  practically  prohibit  the  use  of  circular  rollers  as  the  mini- 
mum diameter  is  specified  as  6"  and  this  size  of  rollers  would  place  them 
so  far  apart  that  the  shoes  would  be  unduly  large.  It  is  not  considered 
good  practice  in  any  case  to  use  rollers  less  than  4"  in  diameter. 

There  is  no  question  but  that  it  is  better  practice  to  use  end  floor 
beams.,  as  the  bottom  ends  of  the  end  posts  are  held  better  than  when  they 
are  omitted  and  the  thermal  expansion  and  contraction  are  better  pro- 
vided for,  as  the  entire  structure  is  supported  upon  the  shoes  and  will 
move  as  a  whole.  In  fact,  there  is  no  excuse  for  omitting  the  end  floor 
beams  other  than  the  slight  saving  in  cost. 

189.  Dead-Load  Stresses  in  Curved  Chord  Pratt  Trusses.— 
As  previously  stated,  the  dead  load  is  considered  as  uniformly  distributed 
over  the  span.     The  approximate  dead  weight  per  foot  of  span  can  be 
obtained  from  (4),  Art.  124,  and  adding  400  Ibs.  to  this  to  provide  for 
the  weight  of  the  deck   (track)  the  total  dead  weight  per  foot  of  span 
for  single-track  bridges  is  obtained,  and  by  dividing  this  by  2  and  mul- 
tiplying by  the  panel  length  the  panel  load  per  truss  is  obtained.     One- 
third  of  this  is  considered  applied  at  the  top  chord  joints  and  two-thirds 
at  the  bottom  chord  joints. 

Let  it  be  required  to  determine  the  dead-load  stresses  in  the  truss 
shown  at  (a),  Fig.  306,  where  P  represents  the  panel  load  per  truss. 

Members  bA  and  AB.  Considering  the  part  of  the  truss  to  the  left 
of  section  1-1,  as  shown  at  (b),  and  resolving  the  forces  vertically  we 
have 


from  which  we  obtain 

Sl=Rsec6 

for  the  stress  in  the  end  post  bA,  and  resolving  the  forces  horizontally 
we  have 

SI  sine  -S2  =  R  sec0  x  sin0  -  S2  =  0, 

from  which  we  obtain 


for  the  stress  in  the  bottom  chord  AB. 

Members  BC,  be  and  bC.     Considering  the  part  of  the  truss  to  the 
left  of  section  2-2,  as  shown  at  (c),  and  taking  moments  about  joint  b, 


416 

we  obtain 


STRUCTURAL  ENGINEERING 


h  h 

for  the  stress  in  bottom  chord  BC.  Next,  resolving  the  stress  S3  (in  top 
chord  be)  into  vertical  and  horizontal  components  at  c  and  taking  moments 
about  C  we  have 


from  which  we  obtain 


Pd 


for  the  horizontal  component  of  the  stress  S3  in  the  top  chord  be,  and 
multiplying  this  by  sec</>  we  obtain 


for  the  stress  in  top  chord  be. 


"    ^  '?   M 


Fig.  306 

Resolving  the  forces  vertically  and  horizontally  and  summing  up  the 
vertical  we  have 

fl-P-F3-F4  =  0, 
from  which  we  obtain 


for  the  vertical  component  of  the  stress  /S4  in  diagonal  bC  and  multiply- 
ing this  by  sec0  we  have 


for  the  stress  in  that  member.     As  (R-P)  is  the  shear  in  panel  BC,  it 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  417 

is  seen  that  the  vertical  component  of  the  stress  in  diagonal  bC  is  equal 
to  the  shear  in  panel  BC  minus  the  vertical  component  of  the  stress  in 
the  top  chord  be.  Another  way  to  obtain  the  stress  in  diagonal  bC  is 
to  prolong  the  top  chord  be  until  it  intersects  the  bottom  chord  at  0  [see 
the  diagram  at  (c)]  and  take  moments  about  O.  Thus,  taking  moments 
about  O  we  have 

Rs-P  (s  +  d)  =  S4:Xt, 
from  which  we  obtain 

S4:  =  RS--P 


for  the  stress  in  the  diagonal  bC. 

In  case  the  last  method  be  used,  the  distances  s  and  t  can  be  deter- 
mined in  the  following  manner:  Triangles  cOC  and  cbm  being  similar 
we  have 

s  +  2d        hi 

d      ~  hl-h 
from  which  we  obtain 

/9*_*1\   ^ 


hl-h 
Similarly,  since  triangles  OnC  and  bBC  are  similar  we  have 


_    _ 

h~~bC~ 
from  which  we  obtain 


_/.  +  2<A 

-  (-be- )  h~ 


Member  cC.    Considering  the  part  of  the  truss  to  the  left  of  section 
3-3,  as  shown  at  (d),  and  resolving  the  forces  vertically  we  obtain 


for  the  stress  in  the  post  cC.     Also,  taking  moments  about  0   [at  (d)] 
we  have 


from  which  we  obtain 


for  the  stress  in  post  cC. 

Members  CD,  cd  and  cD.     Considering  the  part  of  the  truss  to  the 
left  of  section  4-4,  as  shown  at  (e),  and  taking  moments  about  c  we  have 

S7xhl=Rx2d-Pd, 
from  which  we  obtain 

Rx2d    Pd         d 


418  STRUCTURAL  ENGINEERING 

(as  seen  above)  for  the  stress  in  the  bottom  chord  CD.  Next,  resolving 
the  stress  in  the  top  chord  cd  into  horizontal  and  vertical  components  at 
d  and  taking  moments  about  D,  we  obtain 

Rx3d-Px2d-Pxd  d 

~  :6P 


for  the  horizontal  component  and  multiplying  this  by  sec^l   we  obtain 


for  the  stress  in  the  top  chord  cd. 

Resolving  the   forces   horizontally   and  vertically   and   summing  up 
the  vertical  we  have 

R-2P-F8-V9  =  0, 
from  which  we  obtain 

F9  =  R-2P-F8 

for  the  vertical  component  of  the  stress  in  diagonal  cD  and  hence  mul- 
tiplying this  by  sec31  we  obtain 


for  the  stress  in  that  member.     Also,  taking  moments  about  0'  we  obtain 


./ 

Members  dE  and  eD.  Adding  up  from  either  end  of  the  truss  we 
have 

R-3P  =  0 

for  the  shear  in  panel  DE  and  as  the  top  chord  de  is  parallel  to  the 
bottom  DE  it  is  evident  that  the  diagonals  dE  and  eD  have  no  stress 
from  dead  load  and  hence  in  determining  the  dead-load  stresses  in  the 
truss  we  can  ignore  these  members  altogether. 

Members  dD,  DE  and  de.  Considering  the  part  of  the  truss  to  the 
left  of  section  5-5,  as  shown  at  (f),  (ignoring  diagonal  eD)  and  summing- 
up  the  vertical  components  and  forces  we  have 

2 

R~23 
from  which  we  obtain 


for  the  stress  in  post  dD. 

If  R  -2(2/3  P)  -  F8  is  minus  £10  will  be  tension,  as  it  would  have  to 
act  in  the  same  direction  as  R  in  that  case  in  order  that  the  part  of  the 
truss  to  the  left  of  section  5-5  be  in  equilibrium,  while  if  R  -2(2/3  P)  -  FS 
is  plus  £10  will  be  compression,  as  it  would  be  acting  in  the  opposite 
direction  to  that  of  R. 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES  419 

The  stress  in  dD  can  also  be  obtained  by  taking  moments  about  Of 
(see  diagram  at  (f)  ).     Thus,  taking  moments  about  0'  we  obtain 

Rs'  -P(s'  +  d)  +  P(s'  +  2d)  +  f  P(V  4-  3d) 
(,'  +  3«0 

Taking  moments  about  d  we  obtain 


for  the  stress  in  the  bottom  chord  DE,  which,  as  seen  above,  is  equal 
to  H8y  the  horizontal  component  of  the  stress  in  the  top  chord  cd. 

Considering  the  diagonals  eD  and  dE  omitted  from  panel  DE,  as 
there  is  no  dead-load  stress  in  them,  it  is  obvious  that  the  stress  in  top 
chord  de  must  be  equal  and  opposite  to  the  stress  in  the  bottom  chord 
DE  as  these  are  the  only  horizontal  forces  in  the  panel  and  hence  must 
balance  each  other.  So  we  have 

S12  =  S11=SS 

for  the  stress  in  the  top  chord  de. 

Member  bB.  The  only  load  on  the  bridge  that  could  affect  the 
hanger  bB  is  the  one  applied  at  B.  This  load,  as  is  obvious,  is  supported 
directly  by  the  hanger  and  hence  the  dead-load  stress  in  it  is  equal  to  fP. 

As  the  truss  is  symmetrical  about  the  center  of  the  span  and  sym- 
metrically loaded  we  need  not  consider  further  the  dead-load  stresses,  as 
we  have  now  fully  considered  one-half  of  the  span. 

It  is  usual  practice  to  determine  graphically  the  dead-load  stresses 
in  curved  chord  Pratt  trusses. 

190.    Live-Load  Stresses  in  Curved  Chord  Pratt  Trusses.  — 

Chords.     The   criterion    for   the   placing   of   wheel   loads    for   maximum 

stress  in  the  chords  is  exactly  the  same  as  given  in  Art.  91  for  simple 

parallel  chord  trusses.     That  is,  the  maximum  moment  about  any  panel 

point    will    occur    when    the    average 

unit  load  to  the  left  of  the  point  is 

equal  to  the  average  unit  load  on  the 

bridge.      For    example,   to    determine 

the  maximum  live-load  stress  in  chord 

CD    (Fig.    307)    we    would    place    a  Fig.  307 

wheel  at  C  such  that  the  average  unit 

load  to  the  left  of  joint  C  would  be  equal  (as  nearly  as  possible)  to  the 

average  unit  load  on  the  bridge.     Then  by  taking  moments  about  c  of 

the  forces  to  the  left  and  dividing  this  moment  by  h  we  would  obtain 

the  maximum  live-load  stress  in  the  bottom  chord  CD  and  by  multiplying 

this  stress  by  sec<£  we  would  obtain  the  maximum  live-load  stress  in  the 

top  chord  be.     The  live-load  stresses  in  the  other  chord  members  are 

obtained  in  a  similar  manner. 

Web  Members.  Let  it  be  required  to  place  the  wheel  loads  so  that 
maximum  stress  will  occur  in  diagonal  cD  (Fig.  308).  If  the  top  chord 
cd  and  bottom  chord  CD  were  parallel  maximum  stress  would  occur  in 
diagonal  cD  when  the  shear  in  panel  CD  was  a  maximum,  as  the  diagonal 
in  that  case  would  carry  all  of  the  shear  in  the  panel  and  hence  the 


420  STRUCTURAL  ENGINEERING 

criterion  for  maximum  shear  given  in  Art.  90  would  apply;  but  as  the  top 
chord  cd  is  inclined,  and  consequently  carries  some  of  the  shear  in  panel 
CD,  the  shear  criterion  will  not  apply  exactly  and  hence  we  shall  pro- 
ceed to  determine  a  criterion  for  the  placing  of  the  wheels  for  maximum 
stress  in  the  diagonal  cD.  Suppose  the  span  is  loaded  from  H  (Fig. 
308)  up  to  panel  CD  as  shown,  which  is  about  the  same  position  as  for 
maximum  shear  in  panel  CD.  Let  P  be  the  total  load  in  panel  CD,  the 
center  of  gravity  of  which  is  z  distance  from  D,  and  let  W  be  the  total 
load  on  the  bridge,  the  center  of  gravity  of  which  is  x  distance  from  H. 
Considering  the  part  of  the  truss  to  the  left  of  section  1-1  and  taking 
moments  about  0  we  obtain 


for  the  stress  in  diagonal  cD. 

_      Wx 
But  R  =  ~L~ 

Pz 

and  r  -  ~j~  ' 

Substituting  these  values  of  R  and  r  we  obtain 
Ws       Ps        Pa 


Now  suppose  that  there  be  a  slight  movement  of  the  loads  to  the  right 
or  to  the  left,  say,  to  the  left  (W  and  P  remaining  constant),  then  SI, 
x,  and  z  will  each  receive  an  increment,  A*S1,  A.r,  and  A#,  respectively. 
Now  adding  these  increments  in  (1)  we  have 

Ws       Ws      \         Ps        Ps      \       /Pa 

-Li*  +  Ti 

Subtracting  (1)  we  have 


for  the  increment  of  the  stress  in  the  diagonal  cD  due  to  the  slight  move- 
ment of  the  loads.  But  &z  =  A,r,  as  is  obvious.  So  substituting  A#  for  As 
in  (2)  we  have 


Now  Sly  the  stress  in  the  diagonal  cD,  will  be  a  maximum  when 
A$l  =  0.  So  placing  (3)  equals  0  and  reducing  and  transposing  we 
obtain 

TX7          T>  /  ~\ 

ff-?(l+£) <4>- 

Expressing  this  equation  in  words  we  have:  The  average  unit  load 
on  the  bridge  is  equal  to  the  average  unit  load  in  panel  CD  multiplied  by 


DESIGN  OF  SIMPLE  KAILEOAD  BEIDGES 


421 


(1  +  a/s)  when  the  maximum  stress  in  diagonal  cD  occurs.  This  can  be 
taken  as  the  criterion  for  placing  the  loads. 

The  increment  of  the  stress  in  diagonal  cD  (W  remaining  constant) 
can  change  signs  only  when  a  wheel  passes  the  panel  point  D,  so  it  is  seen 
that  a  load  will  be  at  that  point  when  the  maximum  stress  in  the  diagonal 
occurs.  One-half  of  this  load  should  be  considered  in  panel  CD  in  com- 
puting the  value  of  P. 

Next,  let  it  be  required  to  place  the  wheel  loads  so  that  maximum 
stress  will  occur  in  post  cC  (Fig.  308).  Considering  the  part  of  the 


Fig.   308 


truss  to  the  left  of  section  2-2  with  the  wheel  loads  in  about  the  position 
shown  and  taking  moments  about  0'  we  have 


from  which  we  obtain 


Rs' 


-  r 


for  the  stress  in  the  post  cC.     Substituting   (W/L}x,  and   (P/d)z  for 
R  and  r,  respectively,  we  have 


(5). 


Now,  S2,  x,  and  s  will  each  receive  a  small  increment  if  the  loads 
move  slightly  to  the  right  or  left.  Let  A$2,  A,r,  and  A#  be  the  increment, 
respectively,  that  each  receives  due  to  a  slight  movement  to  the  left. 
Then  substituting  these  increments  in  (5)  we  have  • 


d       d 


and  subtracting  (5)  from  this  we  have 

Ws' 


P 

-  -j  As 
d 


£(«'+*) 

and  substituting  A,r  for  As,  since  A,r  =  As,  we  obtain 

Ws'  - 

'~L(s'  +  a)  d 

for  the  increment  of  the  stress  S2  in  the  post  cC  due  to  any  slight  move- 
ment of  the  loads.     Now,  S2  will  be  a  maximum  when  the  increment 


422 


STRUCTURAL  ENGINEERING 


=  0  (see  discussion  of  Art.  90).     So  placing  (6)  equal  to  zero  and 
reducing  we  obtain 

JF_P 
L  ~  d 

which  is  exactly  the  same  as  (4)  except  s'  appears  instead  of  s.  Again, 
suppose  it  be  required  to  place  the  wheel  loads  so  that  maximum  stress 
will  occur  in  diagonals  bC.  By  loading  panel  BC,  very  much  the  same 
as  we  did  CD  (above),  and  taking  moments  about  0'  and  adding  incre- 
ments, in  the  equation  of  moments  we  would  obtain 

JF_P/       o^\ 
L  ~  d\      s' )' 

which  expresses  the  criterion  for  the  placing  of  the  wheel  loads  for 
maximum  stress  in  the  diagonal  bC.  This  last  equation  is  of  exactly 
the  same  form  as  (4)  ;  the  only  difference  is  the  symbols  have  not  the 
same  value  as  they  have  in  (4).  From  this  it  is  seen  that  equation  (4) 
expresses  the  general  criterion  for  the  placing  of  the  wheel  loads  for 
maximum  stress  in  diagonals  and  intermediate  posts. 

In  case  the  chords  are  parallel,  a/s  =  Q  as  *  in  that  case  is  equal  to 
infinity.     The  equation  (4)  then  becomes 

W     P 


Reducing  and  substituting  nd  for  L  we  obtain 


which  is  equation  (5)  of  Art.  90. 

It  is  evident  that  a  live  load  moving  onto  the  bridge  from  the  right 
would  cause  compression  in  diagonal  fE   (Fig.  308)   and  that  this  coin- 


Fig.  309 


pression  would  continue  to  increase  until  the  load  extended  from  H  to  a 
short  distance  beyond  F.  Now  if  this  compression  were  greater  than 
the  dead-load  tension  in  the  diagonal  we  would  have  what  is  known  as  a 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  423 

reversal  of  stress  and  the  diagonal  fE  in  that  case  would  have  to  be 
designed  to  carry  both  tension  and  compression  or  the  member  eF  ,  which 
would  be  known  as  a  counter,  would  have  to  be  inserted  to  carry  the 
reverse  stress.  In  case  the  diagonal  fE  be  composed  of  eye-bars,  which 
are  often  used  in  the  case  of  long  span  bridges,  the  counter  eF  would  be 
used,  as  the  bars  would  not  carry  compression;  but  if  the  diagonal  fE  be 
a  rigid  member  —  that  is,  capable  of  carrying  compression  —  the  counter 
would  be  omitted  and  the  diagonal  would  be  made  sufficient  to  carry 
both  the  tensile  and  compressive  stresses  in  it,  previously  explained  in 
Art.  176  for  diagonal  cD  of  the  150-ft.  span. 

To  determine  the  maximum  live-load  tension  in  the  counter  eF  let 
the  bridge  be  loaded  as  shown  at  (a),  Fig.  309.  Let  P  represent  the 
load  in  the  panel  EF  and  W  the  total  load  on  the  bridge,  and  z  the  dis- 
tance from  F  to  the  center  of  gravity  of  P  and  x  the  distance  from  H  to 
the  center  of  gravity  of  W.  Now,  considering  the  part  of  the  truss  to 
the  left  of  section  6-6  as  shown  at  (b)  and  taking  moments  about  0  we 
have 

R  (L  +  *)  -  r  (a  +  *)  -  St'  -  Slt  =  0, 

from  which  we  obtain 

^K  ..........................  (8) 

for  the  stress  in  the  counter  eF.  Substituting  (W/L)x  and  (P/d)z  for  R 
and  r,  respectively,  we  obtain 


«. 


£1,  as  is  readily  seen,  is  equal  and  opposite  to  the  dead-load  stress  in 
diagonal  fE  and  hence  is  a  known  quantity. 

Now,  adding  increments  to  S,  x,  and  z  in  (9)  and  proceeding  in  the 
same  manner  as  shown  above  in  the  case  of  equations  (1)  and  (5)  we 
obtain 


<»> 


for  the  increment  of  the  stress  in  the  counter  eF.  The  stress  in  the  coun- 
ter will  be  a  maximum  when  this  increment  is  equal  to  zero,  as  previously 
explained.  So  placing  (10)  equal  to  zero  and  transposing  we  obtain 


Expressing  this  in  words  we  have:  The  average  unit  load  on  the 
bridge  is  equal  to  the  average  unit  load  in  the  panel  EF  multiplied  by 
(a  +  s)  -r  (L  +  s)  when  the  maximum  stress  in  the  counter  occurs.  This 
can  be  taken  as  the  criterion  for  placing  the  loads. 

To  determine  the  maximum  stress  in  the  counter  we  would  first 
determine  the  value  of  s,  also  of  t  and  £',  which  can  be  determined  suf- 
ficiently accurately  by  scale  —  in  case  of  counters.  Then  we  would  place 
the  loading  so  as  to  satisfy  equation  (11)  (as  nearly  as  possible)  and 


424 


STRUCTURAL  ENGINEERING 


determine  R  and  r  by  taking  moments  about  H  and  F.  Then  the  stress  S 
in  the  counter  is  readily  found  by  substituting  in  equation  (8).  It  can 
also  be  determined  by  summing  up  the  vertical  forces  and  components 
to  the  left  of  section  6-6.  To  determine  it  in  this  way,  we  would  first 
take  moments  about  F  (Fig.  309)  and  determine  the  horizontal  com- 
ponent of  the  stress  in  the  top  chord  ef.  Then  multiplying  this  by  tan</> 
we  would  have  the  vertical  component  of  the  stress  in  chord  ef,  which  we 
will  designate  as  V\.  Then  by  determining  the  vertical  component  of 
the  dead-load  stress  in  diagonal  fE,  which  we  will  designate  as  V2,  we 
would  have  all  of  the  vertical  forces  to  the  left  of  section  6-6  determined 
except  the  vertical  component  of  the  stress  in  the  counter  eF.  Let  F3 
represent  this  vertical  component.  Then  summing  up  the  vertical  forces 
and  components  to  the  left  of  section  6-6  we  have 


from  which  we  obtain 


for  the  vertical  component  of  the  stress  in  the  counter  eF  and  multiply- 
ing this  by  sec#  we  would  obtain  the  desired  stress  S  in  the  counter.  The 
stress  in  other  counters  is  determined  in  a  like  manner. 

The  maximum  live-load  stress  in  hanger  bB  (Fig.  310)  is  equal  to 
the  maximum  floor  beam  concentration  at  B.  This,  as  explained  in  Arts. 
148  and  171,  is  obtained  by  placing  a  wheel  at  B,  such  that  the  load  in 
panel  AB  will  be  equal  (as  nearly  as  possible)  to  the  load  in  panel  EC. 
After  having  the  loads  thus  placed  the  concentration  at  B,  which  is  equal 
to  the  stress  in  hanger  bB,  is  obtained  by  taking  moments  about  both 
A  and  C.  The  maximum  live-load  stress  in  hanger  gG  is  obtained  in 
the  same  manner  as  for  bB. 

Owing  to  the  top  chord  segments  having  different  slopes  at  the  joints, 
the  intermediate  posts  of  curved  chord  bridges  are  subjected  to  relatively 
greater  tensile  stress  from  live  load  than  the  intermediate  posts  of  parallel 
chord  bridges.  For  example,  let  us  consider  the  case  of  post  cC  (Fig. 


\w      \w  L.ve  Load 

IP       \\9DeodLoad 
\    d    '3 


Fig.  310 

310).  If  panel  points  B  and  C  alone  were  loaded  with  live  load  (ignor- 
ing counter  dC)  it  is  readily  seen  that  post  cC  would  be  in  tension — very 
much  the  same  as  in  the  case  of  parallel  chord  bridges — and  it  is  obvious 
that  this  tensile  stress  would  be  greater  than  it  would  be  if  the  chords 
be  and  cd  had  the  same  slope  at  c,  as  the  chords,  owing  to  their  slopes 
being  different,  pull  upward  (so  to  speak)  on  the  post  cC  and  also  on 
diagonal  cD.  Now,  it  is  evident  that  the  tensile  stress  in  post  cC  will 
increase  as  the  stress  in  the  top  chords  be  and  cd  increases,  and  that  it 


DESIGN  OF  SIMPLE  KAILEOAD  BRIDGES  425 

will  be  a  maximum  when  the  stress  in  diagonal  cD  is  zero,  for  in  that 
case  the  post  alone  would  resist  the  upward  pull  from  the  chords  be 
and  cd.  So  the  problem  in  placing  the  live  load  for  maximum  tension 
in  post  cC,  is  to  place  it  so  as  to  obtain  zero  stress  in  diagonal  cD  ana 
at  the  same  time  as  great  a  stress  as  possible  in  the  top  chords  be  and  cd. 
It  is  customary  in  practice  to  use  an  equivalent  uniform  live  load  (see 
Art.  123)  in  determining  the  tension  in  intermediate  posts  of  curved 
chord  bridges,  as  the  work  is  very  tedious  if  wfyeel  loads  be  used.  So 
we  will  consider  a  uniform  live  load  in  this  case.  Suppose  that  this  live 
load  moves  onto  the  bridge  from  the  left  and  loads  it  from  A  to  m,  just 
so  the  dead-load  tension  in  diagonal  cD  is  reversed.  Then  the  stress  in 
diagonal  cD  and  also  in  counter  dC  would  be  zero.  Now,  as  the  load 
continues  to  move  on  to  the  right  past  m,  the  counter  dC  will  be  in  tension 
and  this  tension  will  increase  steadily,  and  likewise  the  stress  in  the  top 
chords  be  and  cd  (the  stress  in  diagonal  cD  remaining  zero)  until  some 
point  n  is  reached  when  the  counter  dC  will  have  maximum  tension  and 
the  stress  in  diagonal  cD  will  still  be  zero.  Then,  as  the  load  continues 
to  move  on  to  the  right  (past  n),  while  the  stress  in  the  top  chords  be  and 
cd  will  steadily  increase,  the  tension  in  counter  dC  will  steadily  decrease 
until  some  point  o  is  reached  when  the  stress  in  the  counter  dC  is  zero. 
This  position  is  the  one  for  maximum  tension  in  the  post  cC  as  any 
further  movement  of  the  load  to  the  right  would  produce  tension  in  diago- 
nal cD  (which  has  zero  stress  when  the  load  extends  from  A  to  o)  and 
compression  in  post  cC  and  hence  the  tension  in  the  post  would  be  rapidly 
reduced. 

The  determination,  as  to  location,  of  point  o  (the  head  of  the  uniform 
load)  in  any  case  is  very  much  a  matter  of  trial.  We  could  first  load 
from  A  to  D,  for  instance.  Then  compute  the  stress  in  the  counter  dC 
(or  diagonal  cD  in  case  no  counter  is  used)  and  if  a  stress  occurs  with 
the  load  in  that  position  we  could  move  the  load  to  the  right  or  left,  as 
the  case  may  require,  until  the  stress  in  the  counter  is  found  to  be  zero 
and  this  position  of  the  load  is  the  one  required. 

After  the  position  of  the  load  for  maximum  tension  in  the  post  is 
found,  as  Ao,  we  can  take  moments  about  H  and  obtain  R,  the  live-load 
reaction  at  A.  Then  we  can  readily  obtain  the  maximum  live-load  tension 
in  the  post  cC  by  taking  moments  about  0  and  considering  all  of  the  forces 
to  the  left  of  section  2-2.  Thus,  taking  moments  about  0,  we  have 


from  which  we  obtain 

„     W(s  + 

o  = 

for  the  maximum  live-load  tension  in  post  cC,  where  W  —  panel  of  live 
load. 

To  this  tension  should  be  added  the  dead-load  tension  that  occurs  in 
the  post  at  the  same  time.  The  dead-load  tension  in  the  post  is  readily 
obtained  in  the  same  manner  as  shown  above  for  the  live-load  tension. 
Thus,  taking  moments  about  0  we  have 


STRUCTURAL  ENGINEERING 
from  which  we  obtain 

for  the  dead-load  tension  that  occurs  in  the  post  at  the  same  time  the 
maximum  live-load  tension  occurs  in  it.  P  =  panel  of  dead  load  and  R'  = 
total  dead-load  reaction  at  A — the  same  as  occurs  at  H. 

The  maximum  live-load  tension  in  any  other  intermediate  post  is 
obtained  in  a  manner  similar  to  that  shown  above  for  post  cC.  For 
example,  to  obtain  the  maximum  live-load  tension  in  post  dD,  we  would 
load  the  bridge  from  A  on  to  the  right,  until  we  obtained  zero  stress  in 
eD  and  then  we  would  determine  the  maximum  tension  in  the  post  by 
taking  moments  about  0' .  In  the  case  of  post  dD  the  maximum  tension 
would  occur  in  the  member  when  the  span  was  fully  loaded,  for  in  that 
case  both  diagonals,  eD  and  dE,  would  have  zero  stress. 

Complete  Design  of  a  225-Ft.  Single-Track  Through  Pin-Connected 
Curved  Chord  Pratt  Truss  Span 

191.  Data.— 

Length  =  9  panels  @  25'-0"  =  225'-0"  c.c.  end  pins. 

Width  =  17'-0"  c.c.  trusses. 

Height  =  45'-0"  at  center  and  31'-0"  at  hip. 

Stringers  spaced  6'6"  c.c. 

Live  load,  Cooper's  £50  loading. 

Specifications,  A.  R.  E.  Association. 

192.  Design  of  25-Ft.  Stringers. — Proceeding  in  the  manner  out- 
lined in  Art.  170  we  obtain  the  following  for  the  stringers: 

Maximum  End  Shear :  Maximum  Moment : 

Z>  =      5,000  Ibs.  D=    375,000  in.  Ibs. 

L=    71,000  Ibs.  L  =  4,575,000  in.  Ibs. 

1=    66,000  Ibs.  7  =  4,223,000  in.  Ibs. 

142,000  Ibs.          •  9,173,000  in.  Ibs. 

142,000  -r  7,888  =  18n"  9,173,000  -f  45  =  204,000# 

1— web48"xf"  =  18n"  204,000 -f  16,000  =    12.75n" 

End  stiff— Ls3£"x34x£"  2— Ls  6"  x  6"  x  4"  =    10.50°" 

Int.  stiff— Ls  3|"  x  3|  x  |"  J  area  of  web  =      2.25n" 

12.75n" 

193.  Design  of  the  Intermediate  Floor  Beams.—  Proceeding  in 
the  manner  outlined  in  Art.  171  we  obtain  the  following  for  the  inter- 
mediate floor  beams: 

Maximum  End  Shear  Maximum  Moment 

D~    12,000  Ibs.  D=      721,000  in.  Ibs. 

7,=    94,500  Ibs.  L=    5,953,000  in.  Ibs. 

7  =    81,000  Ibs.  7=    5,1 04,000  in.  Ib8. 

187,500  Ibs.  11,778,000  in.  Ibs. 


DESIGN  OF  SIMPLE  KAILEOAD  BRIDGES  427 

187,500  -r  7,644  =  24.5n"  11,778,000  +  53  =  222,000* 

1— web56"xTy  =  24.5n"  222,000 -r  16,000=   13.87n" 

2— Ls  6"  x  6"  x  £"  =   10.50°" 
4  area  of  web  =      3.06n" 


13.56n" 

194.  Design  of  End  Floor  Beams.—  Proceeding  in  the  manner 
outlined  in  Art.  172  and  assuming  the  length  of  each  beam  to  be  14'-6". 
as  they  will  rest  upon  the  shoes  in  this  case,  we  obtain  the  following  for 
the  end  floor  beams: 

Maximum  End  Shear  :  Maximum  Moments  : 

D=      6,500  Ibs.  D=    305,000  in.  Ibs. 

L=    71,000  Ibs.  L  =  3,408,000  in.  Ibs. 

1=    66,000  Ibs.  I  =  3,168,000  in.  Ibs. 

143,500  Ibs.  6,881,000  in.  Ibs. 

143,500  -r  8,266  =  17.37n"  6,881,000  -r  52.37  =  131,000# 

1—  web  56"  x  f"  =  21.00°"  131,000  *  16,000  =      8.19n" 

"  =  7.22-0.75=     6.47n" 


_ 

-  --(12,00 


9.09n" 

195.  Determination  of  Dead-Load  Stresses  in  Trusses.—  From 
(4),  Art.  124,  we  have 

p  =  7x225  +  660  =  2,235  Ibs. 

for  the  approximate  weight  of  metal  per  ft.  of  span  and  adding  400  Ibs. 
for  the  weight  of  the  deck,  we  have 

2,235  +  400  =  2,635  Ibs. 

for  the  total  assumed  dead  load  per  ft.  of  span  or  2,635  4-  2  =  1,318  Ibs.  per 
ft.  of  truss. 

Multiplying  this  by  25,  the  panel  length  in  feet,  we  have 

W  =  1,318  x  25  =  32,950,  say,  33,000  Ibs. 

for  the  panel  load  on  each  truss.  One-third  of  this  will  be  considered  at 
the  top  joints  and  two-thirds  at  the  bottom  joints.  The  dead-load  stresses 
can  be  determined  most  readily  by  graphics. 

Before  we  can  proceed  farther  it  is  necessary  that  we  determine  the 
exact  outline  of  the  truss.  The  panel  points  of  the  top  chord  will  be  on 
the  arc  of  a  parabola.  The  height  of  the  truss  at  the  hip  is  31  ft.  and 
45  ft.  at  the  center  of  the  span,  making  a  drop  of  14  ft.  from  the  center  of 
the  span  to  the  hip.  Laying  off  the  panel  lengths  along  the  bottom  chord 
LO-D  (Fig.  311)  and  taking  C  as  the  vertex  of  the  parabola,  CD  as  the 
,r-axis  and  considering  half  panel  lengths  we  have 

i  *!. 

14  ~  72 
from  which  we  obtain 

x  =  ~x  1  =  0.2857  ft. 


428 


STRUCTURAL  ENGINEERING 


for  the  vertical  distance    from   the  horizontal  line  through   C  down  to 
panel  point  t/4. 

Similarly,  we  have 

fl     §1 

14"  72 
from  which  we  obtain 

*'=j|x  9  =  2.5713  ft. 
for  the  vertical  distance  down  to  panel  point  U3,  and  similarly  we  have 


from  which  we  obtain 


14 


=  7.1428  ft. 


for  the  vertical  distance  down  to  panel  point  U2,  and  thus  we  have  all  of 
the  panel  points  or  joints  of  the  top  chord  located  and  the  outline  of  the 


Fig.  311 

truss  can  be  drawn  as  shown  (Fig.  311).  After  the  outline  of  the  truss  is 
carefully  drawn  (say  j%  scale),  the  dead-load  stress  in  each  member 
can  be  graphically  determined  as  shown  in  Fig.  311;  one-third  of  a  panel 
load  of  dead  load,  which  is  33,000  xj  =  ll,000#,  being  considered  as 
applied  at  each  top  joint,  and  two-thirds,  which  is  33,000  xf  =  22,000#, 
at  each  bottom  joint. 

The  reaction  =  4  x  33,000  =  132,000#,  which  is  applied  at  LO.     We 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


429 


obtain  the  diagram  of  the  stresses  shown  (Fig.  311)  by  first  laying  off  1-2 
(to,  say,  a  ^  scale)  equal  to  the  reaction  and  passing  around  joint  LO 
counter  clock- wise  we  obtain  the  polygon  1-2-3-1  for  joint  LO.  Then 
starting  with  LO-L1,  at  LI,  and  passing  around  joint  LI  counter  clock- 
wise we  obtain  the  polygon  3-2-4-5-3.  Considering  £71  and  passing  around 
counter  clock- wise,  beginning  with  the  11,000#  load,  we  obtain  the  dia- 
gram 6-1-3-5-7-6,  and  so  on  for  the  other  joints  as  fully  explained  in  Ex- 
ample 2,  Art.  96. 

196.  Determination  of  Live-Load  Stresses  and  Impact  in  the 
Trusses. —  In  beginning  the  work  of  determining  the  live-load  stresses 
in  the  truss,  a  diagram  showing  the  entire  truss  should  be  drawn  carefully 
to  scale  (say,  to  a  -fa  scale)  and  upon  this  diagram  the  important  lengths 
and  the  calculated  values  of  the  different  angles  should  be  given  as  shown 
in  Fig.  312. 

End  Post  bA.  The  maximum  stress  will  occur  in  this  member  when 
the  loading  is  placed  for  maximum  shear  in  panel  AB.  By  placing  wheel 


Fig.  312 

4  at  B  (using  Table  A)  we  obtain  2,231  Ibs.  for  the  average  unit  load  on 
the  bridge  and,  considering  one-half  of  the  load  at  B  as  being  in  panel 
AB,  we  obtain  2,400  Ibs.  for  the  average  unit  load  in  the  panel  AB.  This 
position  comes  nearest  to  satisfying  the  criterion  of  Art.  90.  So  taking 
moments  about  J  (Fig.  312)  we  obtain  (using  Table  A) 

R  =  [16,364  +  (284  x  109)  + 1092]  ~|~  =  263,100  Ibs. 
for  the  reaction  at  A,  and  taking  moments  about  B  we  obtain 

=  19,200  Ibs. 


for  the  stringer  reaction  at  A.     Then  for  the  maximum  shear  in  panel 
AB  we  have 

S  =  263,100  - 19,200  =  243,900  Ibs. 

for  Cooper's  .E40  loading  and 

243,900  x  55  =  304,875,  say,  305,000  Ibs. 


430  STRUCTURAL  ENGINEERING 

for  Cooper's  ESQ.    Now  multiplying  this  by  sec#  we  have 
305,000  x  1.28  =  390,400,  say,  390,000  Ibs. 

for  the  maximum  live-load  stress  in  end  post  bA. 

The  load  extends  over  practically  the  entire  span  when  this  maximum 
stress  occurs,  and  hence  the  L  in  the  impact  formula  (Art.  125)  will  be 
taken  as  225.  So  we  have 


=  (3wS25)390'°00  =  233'0001bs- 


for  the  maximum  impact  stress  in  bA. 

Now  adding  the  above  live-load  stress  and  impact  and  the  dead-load 
stress,  given  in  Fig.  311,  together  we  have 

390,000  +  223,000  + 169,000  -  782,000  Ibs. 
for  the  total  maximum  stress  in  the  end  post  bA. 


225' 


Fig.   313 


Diagonal  bC.     By  placing  wheel  3  at  C   (Fig.  313)   and  applying 
Formula  4  of  Art.  190  we  obtain 


By  placing  wheel  2  at  C  we  obtain 
432 


225 

From  this  it  is  seen  that  the  criterion  for  placing  the  wheels  for  maxi- 
mum stress  in  diagonal  bC  is  nearest  satisfied  when  wheel  3  is  at  C.  Then 
placing  wheel  3  at  C  as  shown  in  Fig.  313  and  taking  moments  about  J  we 
obtain  ^  **„. 

R  =  (16,364  +  22,436  +  79*)  -£—  =  200,180  Ibs. 


for  the  reaction  at  A.     Then  taking  moments  about  C  of  the  forces  to  left 
we  obtain 

HI  =  [ (200,180  x  50)  -  230,000]  -r  37.86  =  258,290  Ibs. 

for  the  horizontal  component  of  the  stress  in  the  top  chord  be  and  multi- 
plying this  by  tan<£  we  have 

V\  =  258,290  x  0.2744  =  70,875  Ibs. 

for  the  vertical  component  of  the  stress  in  the  top  chord  be. 
Taking  moments  about  C  of  wheels  1  and  2  we  obtain 

r  =  9,200  Ibs. 


DESIGN  OF  SIMPLE  BAILKOAD  BRIDGES 


431 


Now  adding  up  all  of  the  vertical  forces  and  components  to  the  left 
of  sections  2-2  we  have 

from  which  we  obtain 

V  =  R-r-Vl 

and  substituting  in  the  numerical  values  given  above  we  have 

V  =  200,180  -  9,200  -  70,875  =  120,105  Ibs. 

for  the  vertical  component  of  the  stress  in  diagonal  bC  and  multiplying 
this  by  sec0  we  obtain 

S  =  120,105  x  1.28  =  153,734  Ibs. 

for  the  stress  in  the  diagonal  due  to  Cooper's  J240  and  multiplying  this 
by  50/40  we  obtain  192,167,  say  192,000  Ibs.  for  the  maximum  live-load 
stress  in  diagonal  bC. 

When  this  maximum  stress  occurs  the  span  is  loaded  practically  from 
C  to  J  and  hence  the  L  in  the  impact  formula  will  be  taken  as  175  ft.  So 
we  have 

1  = 


for  the  impact  stress  in  diagonal  bC. 

Now  adding  together  the  above  live-load  stress  and  impact  and  the 
dead-load  stress,  given  in  Fig.  311,  we  have 

192,000  + 121,000  +  73,000  =  386,000  Ibs. 

for  the  total  maximum  tension  in  diagonal  bC. 

Intermediate  Post  cC.     Placing  wheel  3  at  D  and  applying  Formula 
4  of  Art.  190  we  obtain 


and  placing  wheel  2  at  D  we  obtain 

382  _ 
225" 


Fig.  314 

From  this  it  is  seen  that  the  criterion  for  placing  the  wheel  loads  for 
maximum  stress  in  post  cC  is  nearest  satisfied  when  wheel  2  is  at  D. 
Then  placing  wheel  2  at  D,  as  shown  in  Fig.  314,  and  taking  moments 
about  J  we  obtain 

R  =  145,200  Ibs. 


432  STRUCTURAL  ENGINEERING 

for  the  reaction  at  A  and  taking  moments  about  D  of  wheel  1  we  obtain 

r  =  (10,000  x  8)  -f  25  =  3,200  Ibs. 

for  the  floor  beam  concentration  at  C.  Then  taking  moments  abouc  C 
of  all  the  forces  and  components  to  the  left  of  section  3-3  we  obtain 

H=(Rx5Q)+h, 

and  substituting  in  the  numerical  values  given  above  we  have 
H  =  (145,200  x  50)  -  37.86  =  191,700  Ibs. 

for  the  horizontal  component  of  the  stress  in  the  top  chord  be.  Then 
multiplying  this  component  by  tan<£  we  obtain 

F=  191,700  x  0.2744  =  52,600  Ibs., 

for  the  vertical  component  of  the  stress  in  the  top  chord  be.  Now  adding 
up  (algebraically)  all  of  the  vertical  forces  and  components  to  the  left  of 
section  3-3  we  have 

R-r-F-S  =  Q, 

and  substituting  in  the  numerical  values  given  above  and  transposing 
we  obtain 

S  =  145,200  -  3,200  -  52,600  =  89,400  Ibs. 

for  the  stress  in  post  cC  due  to  Cooper's  £40,  and  multiplying  this  by 
50/40  we  obtain  111,700,  say,  112,000  Ibs.  for  the  maximum  live-load 
compression  in  post  cC. 

When  this  maximum  stress  occurs  the  span  is  loaded  practically  from 
D  to  J  and  hence  the  L  in  the  impact  formula  will  be  taken  as  150.  So 
we  have 


/=  112>°°°  =  ™>0QQ  lbs- 


for  the  impact  stress  in  post  cC.    Now  adding  together  the  above  live-load 
stress  and  impact  and  the  dead-load  stress,  given  in  Fig.  311,  we  have 

112,000  +  75,000  +  34,000  =  221,000  lbs. 

for  the  total  maximum  compression  in  post  cC. 

Diagonal  cD.  Placing  wheel  3  at  D  (Fig.  312)  and  applying  equa- 
tion 4  of  Art.  190  we  obtain 

392  = 
225  ~ 

This  position,  as  can  be  verified  by  trial,  comes  nearest  to  satisfying 
the  criterion  for  maximum  stress  in  diagonal  cD.  So  placing  wheel  3  at 
D,  as  shown  in  Fig.  315,  and  taking  moments  about  J  we  obtain 


R=  (16,364  +  15,336  -f542)-  153,700  lbs. 


for  the  reaction  at  A.    Taking  moments  about  D  of  the  wheels  to  the  left 
we  obtain 


DESIGN  OF  SIMPLE  EAILKOAD  BRIDGES 


433 


for  the  floor  beam  concentration  at  C.     Next,  taking  moments  about  D  of 
the  forces  and  components  to  the  left  of  section  4-4  we  obtain 

Hl     (153,700x50)     (9,200  x25) 


r  4 


I 
54' 


L  = 


Figr.   315 

for  the  horizontal  component  of  the  stress  in  the  top  chord  cd.     Multiply- 
ing this  by  tan<^l  we  obtain 

Vl^Hlx  tan<£l  =  266,200  x  0.1828  =  48.660  Ibs. 

for  the  vertical  component  of  the  stress  in  top  chord  cd.    Now  summing  up 
the  vertical  forces  and  components  to  the  left  of  section  4-4  we  have 


Substituting  the  numerical  values  given  above  and  transposing  we  obtain 
V=  153,700  -  9,200  -  48,660  =  95,840  Ibs. 

for  the  vertical  component  of  the  stress  in  diagonal  cD  and  multiplying 
this  by  secfll  we  have 

S  =  V  sec(91  =  95,840  x  1.199  =  114,912  Ibs. 

for  the  stress  in  diagonal  cD  due  to  Cooper's  £40  loading.  Then  by  mul- 
tiplying this  by  50/40  we  obtain  143,635,  say,  144,000  Ibs.  for  the*  maxi- 
mum tension  in  diagonal  cD  due  to  the  -E50  loading. 

When  this  maximum  stress  occurs  the  span  is  practically  loaded  from 
D  to  J  and  hence  the  L  in  the  impact  formula  will  be  taken  as  150  ft.  So 
we  have 


for  the  impact  stress  in  diagonal  cD. 

Now  adding  together  the  above  live-load  stress  and  impact  and  the 
dead-load  stress,  given  in  Fig.  311,  we  obtain 

144,000  +  96,000  +  40,000  =  280,000  Ibs. 

for  the  total  maximum  tension  in  diagonal  cD. 

Intermediate  Post  dD.    Placing  wheel  2  at  E  (Fig.  312)  and  apply- 
ing equation  4  of  Art.  190  we  obtain 

332  =__,      ,20  A       75 
225" 


434 


STRUCTURAL  ENGINEERING 


This  position  of  the  wheels  practically  satisfies  the  criterion  for  maxi- 
mum stress  in  post  dD.  So  placing  wheel  2  at  E,  as  shown  in  Fig.  316, 
and  taking  moments  about  J  we  obtain  (using  Table  A) 

R  =  105,500  Ibs. 


Fig.   316 

for  the  reaction  at  Ay  and  taking  moments  about  E  of  wheel  1  we  obtain 

r  =  3,200  Ibs. 

for  the  floor  beam  concentration  at  D.  Then  taking  moments  about  D 
of  the  forces  and  components  to  the  left  of  section  5-5  we  obtain 

HI  =  186,500  Ibs. 

for  the  horizontal  component  of  the  stress  in  top  chord  cd  and  multiply- 
ing this  by  tan</>!  we  obtain 

VI  =  186,500  x  0.1828  =  34,092  Ibs. 

for  the  vertical  components  of  the  stress  in  top  chord  cd. 

Now,  by  summing  up  (algebraically)  the  vertical  forces  to  the  left 
of  section  5-5  we  obtain 

S  =  105,500  -  3,200  -  34,092  =  68,208  Ibs. 

for  the  stress  in  post  dD  due  to  Cooper's  £40  and  multiplying  this  by 
50/40  we  obtain  85,260,  say,  85,000  Ibs.  for  the  maximum  compression  in 
post  dD  due  to  the  E50  loading. 

When  this  maximum  stress  occurs  the  span  is  practically  loaded  from 
E  to  J  and  hence  the  L  in  the  impact  formula  will  be  taken  as  125  ft.  So 
we  have 


for  the  impact  stress  in  post  dD. 

Now,  adding  together  the  above  live-load  stress  and  impact,  and  the 
dead-load  stress,  given  in  Fig.  311,  we  obtain 

85,000  +  60,000  +  12,000  =  157,000  Ibs. 

for  the  total  maximum  compression  in  post  dD. 

Diagonal  dE.     Placing  wheel  3  at  E  (Fig.  312)  we  obtain 


342  ,40 

_  =1.52  and  _ 


75 


DESIGN  OF  SIMPLE  RAILROAD  BEIDGES 


435 


This  position  of  the  load  comes  nearest  to  satisfying  the  criterion  for 
maximum  stress  in  diagonal  dE.  So  placing  wheel  3  at  E,  as  shown  in 
Fig.  317,  and  taking  moments  about  J  we  obtain 

R  =  113,200  Ibs. 

for  the  reaction  at  A,  and  taking  moments  about  E  of  the  wheels  to  the 
left  we  obtain 

r  =  9,200  Ibs. 

for  the  floor  beam  concentration  at  D.     Then  taking  moments  about  E 
of  the  forces  and  components  to  the  left  we  obtain 

HI  =  248,100  Ibs. 

for  the  horizontal  component  of  the  stress  in  the  top  chord  de  and  mul- 
tiplying this  by  tan<£2  we  obtain 

VI  =  248,100  x  0.0912  =  22,626  Ibs. 

for  the  vertical  component  of  the  stress  in  top  chord  de. 

Now,  summing  up  the  vertical  forces  and  components  to  the  left  of 
section  6-6  we  obtain 

F=  113,200  -  9,200  -  22,626  =  81,374  Ibs. 

for  the  vertical  component  of  the  stress  in  diagonal  dE,  and  multiplying 
this  by  sec$2  we  obtain  94,393  Ibs.  for  the  stress  in  diagonal  dE  due  to 


Fig.   317 


£40  loading;  and  multiplying  this  by  50/40  we  obtain  117,991,  say,  118,- 
000  Ibs.  for  the  maximum  live-load  stress  in  the  member.  When  this  maxi- 
mum stress  occurs,  the  span  is  loaded  practically  from  E  to  J  and  hence 
the  L  in  the  impact  formula  will  be  taken  as  125  ft.  So  we  have 


/= 


118>000  =  83>soo>  say>  83>000  lbs- 


for  the  impact  stress  in  diagonal  dE. 

Now,  adding  together  the  above  live-load  stress  and  impact  and  the 
dead-load  stress  given  in  Fig.  311,  we  obtain 

118,000  +  83,000  +  21,000  =  222,000  lbs. 
for  the  total  maximum  stress  in  diagonal  dE.  , 


436  STRUCTURAL  ENGINEERING 

Intermediate  Post  eE.     Placing  wheel  2  at  F  (Fig.  312)  we  obtain 


and  placing  wheel  3  at  F  we  obtain 

292      1  Q       ,  40  /, 
—  =  1.3  and -^1 

From  this  it  is  seen  that  the  criterion  (Art.  190)  for  maximum  stress 
in  post  eE  is  the  nearest  satisfied  with  wheel  2  at  F.  So  placing  wheel  2 
at  F,  as  shown  in  Fig.  318,  and  taking  moments  about  J  we  obtain 
(using  Table  A) 

R  =  71,466  Ibs. 


Fig.  318 

for  the  reaction  at  A  and  taking  moments  about  F  of  the  wheel  to  the 
left  we  obtain 

r=  3,200  Ibs. 

for  the  floor  beam  concentration  at  E.  Then  taking  moments  about  E 
of  the  forces  and  components  to  the  left  of  section  7-7  we  obtain 

HI  =  159,843  Ibs. 

for  the  horizontal  component  of  the  stress  in  the  top  chord  de  and  multi- 
plying this  by  tan<£2  we  obtain 

Fl  =  14,577  Ibs. 

for  the  vertical  component  of  the  stress  in  top  chord  de.  Then  adding 
algebraically  the  vertical  forces  and  components  to  the  left  of  section 
7-7  we  obtain 

S  =  71,466  -  3,200  -  14,577  =  53,689  Ibs. 

for  the  maximum  compression  in  post  eE  due  to  the  E40  loading  and  mul- 
tiplying this  by  50/40  we  obtain  67,111,  say,  67,000  Ibs.  for  the  maximum 
live-load  compression  in  post  eE. 

When  this  maximum  stress  occurs  the  span  is  loaded  practically 
from  F  to  J  and  hence  the  L  in  the  impact  formula  will  be  taken  as  100. 
So  we  have 


I  =  ( 


67,000  =  50,250,  say,  50,000  Ibs. 


for  the  impact  stress  in  post  eE. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


437 


Now,  adding  the  above  live-load  stress  and  impact,  and  the  dead- 
load  stress  given  in  Fig.  311  we  obtain 

67,000  +  50,000  -  5,000  =  112,000  Ibs. 

for  the  total  maximum  compression  in  post  eE. 

Diagonal  eF.  As  the  top  chord  ef  (Fig.  312)  is  parallel  to  bottom 
chord  EF  the  diagonal  eF  will  carry  all  of  the  maximum  shear  in  panel 
EF  when  the  load  moves  onto  the  bridge  from  the  right  and  hence  the 
ordinary  criterion  of  Art.  90  for  the  maximum  shear  in  panel  EF  applies. 
According  to  this  criterion,  the  maximum  shear  will  occur  in  panel  EF 
when  the  average  unit  load  on  the  bridge  is  equal  to  the  average  unit  load 
in  the  panel. 

This  is  the  nearest  satisfied  when  wheel  3  is  at  F.  So  placing  wheel 
3  at  F  as  shown  in  Fig.  319  and  taking  moments  about  J  we  obtain 

R  =  77,851  Ibs. 


Fig.  319 

for  the  reaction  at  A  and  taking  moments  about  F  of  the  wheels  to  the 
left  we  obtain 

r  =  9,200  Ibs. 

for  the  floor  beam  concentration  at  E.  Then  for  the  maximum  shear  in 
panel  EF  we  have 

(R  -  r)  =  (77,851  -  9,200)  =  68,651  Ibs. 

This  is  assumed  to  be  carried  altogether  by  the  diagonal  eF,  so,  evi- 
dently the  vertical  component  of  the  stress  in  that  member  must  be  equal 
to  this  shear;  that  is, 

F  =  68,651  Ibs. 

Then  multiplying  this  by  sec03  we  obtain 

S  =  68,652  x  1.145  =  78,605  Ibs. 

for  the  stress  in  the  diagonal  eF  due  to  £40  loading  and  multiplying  this 
by  50/40  we  obtain  98,256,  say  98,000  Ibs.  for  the  maximum  live-load 
stress  in  diagonal  eF. 

When  this  maximum  stress  occurs  the  span  is  loaded  practically  from 
F  to  J  and  hence  the  L  in  the  impact  formula  will  be  taken  as  100. 

Then  we  have 


/=! 


\100  +  300 
for  the  impact  in  diagonal  eF. 


^ — \  98,000  =  73,500,  say  73,000  Ibs. 
J  -f  300  / 


438 


STRUCTURAL  ENGINEERING 


Now,  adding  the  above  live-load  stress  and  impact,  and  the  dead-load 
stress  given  in  Fig.  311  we  obtain 

98,000  +  73,000  +  00,000  =  171,000  Ibs. 
for  the  total  maximum  stress  in  diagonal  eF. 


Fig.  320 

Counter  eD.  Placing  wheel  3  at  D  as  shown  at  (a),  Fig.  320,  and 
applying  Formula  11  of  Art.  190  we  have  (not  considering  wheel  15  on 
the  bridge) 

232      inq   nd40  -  ?fi 

—      1.03  and  -  =  1.26. 


This  position  of  the  wheels  comes  nearest  to  satisfying  the  criterion 
for  maximum  stress  in  counter  eD.  So  placing  wheel  3  at  D  and  taking 
moments  about  A  we  obtain 


R  =  (10,816)  =  48,071  Ibs. 


for  the  reaction  at  J  and  taking  moments  about  D  of  the  wheels  to  the 
right  we  obtain 

r  =  9,200  Ibs. 

for  the  floor  beam  concentration  at  E. 

SI  is  equal  and  opposite  to  the  dead-load  stress  in  diagonal  dE  which 
is  given  in  Fig.  311  as  21,000  Ibs.  Then,  resolving  SI  into  horizontal 
and  vertical  components  at  E,  we  obtain 

VI  =  SI  -f  sec02  =  21,000  -=-  1.16  =  18,103  Ibs. 

for  the  vertical  component  of  the  stress  in  diagonal  dE. 

Taking  moments  about  D  of  the  forces  and  components  to  the  right 
we  obtain 

(R  x  150)  -  (VI  x  25)  -  (r  x  25)  -  (H2  x  42.43)  =  0. 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  439 

Substituting  in  the  numerical  values  given  above  and  transposing  and 
reducing  we  obtain 

H2  =  153,853  Ibs. 

for  the  horizontal  component  of  the  stress  in  top  chord  de  and  multiply  • 
ing  this  by  tan</>2  we  obtain 

F2  =  153,853  x  0.0912  =  14,031  Ibs. 

for  the  vertical  component  of  the  stress  in  top  chord  de. 

Now,  summing  up  all  the  vertical  forces  and  components  to  the 
right  of  section  9-9  we  obtain 

and  substituting  in  the  numerical  values  given  above  we  have 
V  =  48,071  -  9,200  - 18,103  + 14,031  =  34,799  Ibs. 

for  the  vertical  component  of  the  stress  in  counter  eD  and  multiplying 
this  by  sec#3  we  have 

S  =  34,799  x  1.145  =  39,844  Ibs. 

for  the  maximum  stress  in  counter  eD  due  to  the  £40  loading  and  mul- 
tiplying this  by  50/40  we  obtain  49,805,  say,  50,000  Ibs.  for  the  maximum 
live-load  stress  in  counter  eD  due  to  J550  loading. 

When  this  maximum  stress  occurs  the  span  is  loaded  practically  from 
D  to  A  and  hence  the  L  in  the  impact  formula  will  be  taken  as  75. 

Then  we  have 

1  = 

for  the  impact  stress  in  counter  eD. 

Now,  adding  the  above  live-load  stress  and  impact  together  we  have 
(counters  carry  no  dead  load) 

50,000  +  40,000  =  90,000  Ibs. 

for  the  total  maximum  stress  in  counter  eD. 

No  counter  is  needed  in  panel  CD  as  the  live-load  shear  is  not  suf- 
ficient to  reverse  the  dead-load  shear;  or,  in  other  words,  the  dead-load 
tension  in  diagonal  cD  is  greater  than  the  live-load  compression  in  it. 

Tension  in  Post  dD.  The  maximum  shear  in  end  panel  AB  was 
found  above  (in  determining  the  stress  in  end  post  bA)  to  be  305,000  Ibs. 

Substituting  this  shear  in  the  formula  given  in  Art.  123  we  obtain 


for  the  equivalent  uniform  live  load  per  foot  of  truss  for  determining  the 
stress  in  the  web  members  and  hence  we  can  use  this  load  for  determin- 
ing the  tension  in  intermediate  posts.  Then  for  the  panel  load  we  have 

W  =  3,050  x  25  =  76,250,  say  76,000  Ibs. 

The  first  part  of  our  problem  now  is  to  place  this  uniform  load  so 
as  to  obtain  zero  stress  in  counter  eD  and  also  in  diagonal  dE  and  at  the 


440 


STRUCTURAL  ENGINEERING 


same  time  as  great  a  stress  in  chords  cd  and  de  as  possible  (see  Art.  190). 
The  equivalent  uniform  live  load  will  produce  practically  the  same 
stress  in  the  counter  eD  (Fig.  320)  as  the  wheel  loads.  So,  if  panel 
points  B,  C,  and  D  are  loaded  with  the  above  uniform  load  the  counter 
eD  will  have  a  maximum  tensile  stress  of  50,000  Ibs.,  the  same  as  pre- 
viously found  for  wheel  loads.  Dividing  this  by  sec03  we  obtain 

50,000  4- 1.145  =  44,000  Ibs.  (about) 

for  the  vertical  component  of  the  maximum  tensile  stress  in  counter  eD. 
Now,  any  load  at  E,  F,  or  any  panel  point  to  the  right  of  panel  DE  will 
tend  to  reduce  the  tension  in  counter  eD.  What  we  desire  is  to  place  the 
load  just  so  that  the  50,000  Ibs.  tension  in  the  counter  is  exactly  counter- 
acted, for  then  the  counter  eD  and  diagonal  dE  will  have  zero  stress  and 
hence  the  post  dE  will  have  maximum  live-load  tension. 

By  loading  panel  point  E  alone  we  obtain  a  reaction  at  A  of  76,- 
000  x  5/9  =  42,000  Ibs.    Now,  as  the  vertical  component  of  the  50,000  Ibs. 


Fig.  321 


tensile  stress  in  the  counter  eD  alone  is  about  44,000  Ibs.,  it  is  obvious 
that  a  panel  load  at  E  will  not  reverse  the  50,000  Ibs.  tension  in  the 
counter.  So  next,  let  us  place  a  panel  load  at  both  E  and  F.  Then  for 
the  reaction  at  A,  which  is  equal  to  the  negative  shear  in  panel  DE  due 
to  the  loads  at  E  and  F,  we  obtain 

76,000  x^  +  76,000  x  |  =  76,000 Ibs.  (about), 
y  y 

Taking  moments  about  D  (panel  points  E  and  F  alone  being  loaded)  we 
obtain 

HI  =  (76,000  x  75)  -r 42.43  =  135,000  Ibs.  (about) 

for  the  horizontal  component  of  the  stress  in  top  chord  de.  Then  mul- 
tiplying this  by  tan</>2  we  obtain 

VI  =  135,000  x  0.0912  =  12,300  Ibs.  (about) 

for  the  vertical  component  of  the  stress  in  top  chord  de  due  to  the  panel 
loads  at  E  and  F. 


DESIGN  OF  SIMPLE  KAILKOAD  BRIDGES  441 

Then  subtracting  this  from  the  reaction  we  obtain 
76,000  -  12,300  =  63,700  Ibs. 

for  the  vertical  component  in  counter  eD.  This  is  too  great,  as  the  coun- 
ter would  be  more  than  reversed  and  hence  the  diagonal  dE  would  then 
be  in  tension. 

Next,  suppose  the  load  extends  just  up  to  panel  point  F  so  that  the 
load  at  F  will  be  only  a  half  panel  load  or  38,000  Ibs.  Then  for  the 
reaction  at  A  we  have 

76,000  x  |  +  38,000  x  |=  59,000  Ibs.  (about). 
y  y 

Now,  taking  moments  about  D  we  obtain 

0.0912  =  9,500  Ibs.  (about) 

for  the  vertical  component  of  the  stress  in  the  top  chord  de. 
Then  subtracting  this  from  the  reaction  at  A  we  have 

59,000  -9,500  =  49,500  Ibs. 

for  the  vertical  component  of  the  compressive  stress  in  counter  eD.  As  is 
seen,  this  is  too  large.  So  let  us  extend  the  load  just  20  ft.  beyond  panel 
point  E  as  shown.  Then  the  load  at  E  will  be  75,000  Ibs.  and  at  F  it 
will  be  24,000  Ibs.  The  reaction  at  A  due  to  these  two  loads  will  be 


(  75,000  x^\  +  ^24,000  x|U  52,300  Ibs.  (about). 

btain 
8^400  ibs.  (about) 


Then  taking  moments  about  D  we  obtain 
/52,300x75\ 


\      42.43      / 

for  the  vertical  component  of  the  stress  in  the  top  chord  de.  Now,  sub- 
tracting this  from  the  reaction  at  A  we  obtain 

52,300 -8,400  =  43,900  Ibs. 

for  the  vertical  component  of  the  compressive  stress  in  the  counter  eD 
due  to  the  loads  at  E  and  F  which  just  reverses  the  maximum  tension 
in  the  counter.  So  this  is  the  position  of  the  load  being  sought  for.  That 
is,  by  placing  the  load  so  that  it  extends  from  A  to  20  ft.  beyond  E  the 
stress  in  both  the  counter  eD  and  diagonal  dE  will  be  zero  and  the  chords 
cd  and  de  will  have  the  greatest  stress  possible  with  that  condition  and 
hence  the  tension  in  post  dD  will  be  a  maximum. 

So  placing  the  load  in  that  position  and  taking  moments  about  A 
we  obtain 

R'  =  (3,050  x  120)~=  97,644  Ibs. 


.  (about) 


for  the  reaction  at  J. 

Then  taking  moments  about  D  we  obtain 


442  STRUCTURAL  ENGINEERING 

for  the  horizontal  component  of  the  stress  in  top  chord  de  and  multiplying 
this  by  sec<£2  we  obtain 

418,000  x  1.004  =  420,000  Ibs.  (about) 

for  the  stress  in  top  chord  de.  Now  as  the  stress  in  diagonal  dE  is  zero, 
the  tensile  stress  in  the  post  dD  can  be  obtained  very  quickly  by  drawing 
the  diagram  shown  at  (b).  From  this  diagram  the  tension  in  post  dD 
is  found  to  be  about  38,000  Ibs. 

As  is  seen,  when  this  maximum  tension  occurs  in  the  post  dD  there 
is  120  ft.  of  load  on  the  bridge,  so  we  have 


1  =     lon^Qnn    38>000  =  26>000  lbs'  (about) 


for  the  impact  stress. 

Now,  assuming  diagonal  dE  as  having  zero  stress  we  can  obtain 
the  dead-load  tension  in  post  dD  by  drawing  the  diagram  shown  at  (d)0 

The  stress  in  top  chord  cd  is  given  in  Fig.  311  as  177,000  lbs.  Then 
drawing  1-3  equal  (by  scale)  to  177,000  and  parallel  to  cd,  and  1-2  and 
3-2  parallel  to  de  and  dD,  respectively,  we  have  the  dead-load  tension 
in  the  post  given  by  the  line  3-2.  This  tension  is  found  in  this  manner 
to  be  16,000  lbs. 

Now  adding  together  the  above  live-  and  dead-load  stresses  and 
impact  we  obtain 

38,000  +  26,000  + 16,000  =  80,000  lbs. 

for  the  total  maximum  tension  in  post  dD. 

Tension  in  Post  eE.  As  is  obvious,  the  maximum  tension  will  occur 
in  post  eE  when  the  span  is  fully  loaded,  for  in  that  case  both  of  the 
diagonals  eD  and  fE  have  zero  stress.  The  counter  eD  also  has  zero 
stress  at  the  same  time. 

Now,  as  we  are  using  a  uniform  live  load  we  can  readily  obtain  the 
live-load  tension  in  post  eE  by  direct  proportion  as  it  will  be  propor- 
tional to  the  dead-load  tension  given  in  Fig.  311. 

So,  letting  S  represent  the  live-load  tension,  we  have 

S          3,050 
5,000  "  1,318' 
from  which  we  obtain 

S  =  11,570,  say,  12,000  lbs. 

for  the  maximum  live-load  tension  in  post  eE  where  3,050  is  the  uniform 
live  load  and  1,318  the  dead  load  per  ft.  of  truss. 
For  the  impact  we  have 


Now,  adding  together  the  above  live-load  stress  and  impact,  and  the 
dead-load  stress  given  in  Fig.  311,  we  obtain 

12,000  +  7,000  +  5,000  =  24,000  lbs. 
for  the  total  maximum  tension  in  post  eE. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  443 

Bottom  Chords  AB  and  BC.  The  maximum  stress  can  be  obtained 
in  these  members  by  placing  a  load  at  B  such  that  the  average  unit  load 
to  the  left  is  equal  to  the  average  unit  load  on  the  bridge  (see  Art.  90) 
and  taking  moments  about  b;  but  the  same  can  be  found  by  simply  mul- 
tiplying the  maximum  shear  in  panel  AB  by  tan#.  The  maximum  shear 
was  found  above  (in  determining  the  stress  in  the  end  post  bA)  to  be 
305,000  Ibs.  So  we  have 

305,000  x  0.8066  =  246,013,  say,  246,000  Ibs. 

for  the  live-load  stress  in  each  of  the  bottom  chords  AB  and  BC. 

When  this  maximum  stress  occurs,  the  load  extends  practically  over 
the  entire  span,  so  for  the  impact  stress  in  each  of  these  chords  we  have 


7  =  246'00°  =  140'571>  sa^  141>000  lbs' 


Now,  adding  together  the  above  live-load  stress  and  impact,  and 
the  dead-load  stress,  given  in  Fig.  311,  we  obtain 

246,000  +  141,000  +  106,000  =  493,000  lbs. 

for  the  total  maximum  stress  in  each  of  the  bottom  chords  AB  and  BC. 

Bottom  Chord  CD.  The  stress  in  this  member  (see  Fig.  312)  is 
found  by  taking  moments  about  panel  point  c.  Then,  evidently,  the 
stress  will  be  a  maximum  when  the  load  is  placed  so  that  this  moment 
is  a  maximum.  According  to  Art.  90,  the  moment  will  be  a  maximum 
when  a  load  is  at  C  such  that  the  average  unit  load  to  the  left  is  equal 


>03' 


Fig.   322 

to  the  average  unit  load  on  the  bridge.     Placing  wheel  7  at  C,  as  shown 
in  Fig.  322,  we  have 

(103  +  6.5)^.5?  =  2,190  lbs. 
ot) 

for  the  average  unit  load  to  the  left  of  C  (considering  one-half  of  wheel 
7  as  being  to  the  left  of  C)  and 

[284 +(103x2)]^?  =2,177  lbs. 

for  the  average  unit  load  on  the  bridge.     This  position  of  the  load  comes 
nearest  to  satisfying  the  criterion  for  maximum  moment  about  c  (or  C). 
Then  by  taking  moments  about  J  (using  Table  A)  we  obtain 

R  =  [16,364  +  (284  x  103)  + 103  *]  ^^  =  249,880  lbs.  (about) 

/v/vO 

for  the  reaction  at  A.     Next,  taking  moments  about  c  of  the  forces  to  the 
left  and  dividing  by  the  height  of  the  truss  at  that  point  we  obtain 


444  STRUCTURAL  ENGINEERING 

S  =  [  (249,880  x  50)  -  (2,155  x  1,000)  ]  —  ~  =  273,000  Ibs.  (about) 

OY.OO 

for  the  maximum  stress  in  the  bottom  chord  CD  due  to  the  £40  loading 
and  multiplying  this  by  50/40  we  obtain  341,250,  say  341,000  Ibs.  due  to 
the  £50  loading  which  is  the  maximum  live-load  stress  desired. 

When  this  maximum  stress  occurs  the  live  load  extends  over  practi- 
cally the  entire  span  and  hence  the  L  in  the  impact  formula  will  be  taken 
as  225.  So  for  the  impact  stress  we  have 


Now,  adding  together  the  above  live-load  stress  and  impact,  and  the 
dead-load  stress  given  in  Fig.  311,  we  obtain 

341,000  +  194,000  +  152,000  =  687,000  Ibs. 

for  the  total  maximum  stress  in  bottom  chord  CD. 

Top  Chord  be.  Considering  the  forces  to  the  left  of  section  1-1  (Fig. 
322)  it  is  readily  seen  that  the  horizontal  component  of  the  maximum 
stress  in  top  chord  be  is  equal  to  the  maximum  stress  in  bottom  chord  CD. 
So  by  multiplying  the  stress  in  CD  (found  above)  by  sec<£  (see  Art.  190) 
we  obtain 

SI  =  341,000  x  1.037  -  353,617,  say,  354,000  Ibs. 

for  the  maximum  live-load  stress  in  top  chord  be.     For  the  impact  we 
have 

/  -  (aJ°030Q)  35^000  =  20^000  lbs-  (about)  • 

Now,  adding  together  the  above  live-load  stress  and  impact,  and 
the  dead-load  stress  given  in  Fig.  311,  we  obtain 

354,000  +  202,000  +  157,000  =  713,000  lbs. 

for  the  total  maximum  stress  in  top  chord  be. 

Bottom  Chord  DE.  The  stress  in  this  member  is  found  by  taking 
moments  about  panel  point  d  and  hence  the  stress  in  the  member  will  be 
a  maximum  when  the  moment  about  d  is  a  maximum.  Placing  wheel  11 
at  D,  as  shown  in  Fig.  323,  we  have 


(152  +  10)  =  2,160 

•  o 

for  the  average  unit  load  to  the  left  of  D  (or  d)  and 
(284  +  210)^=2,195 

for  the  average  unit  load  on  the  bridge.  This  position  of  the  load  come* 
the  nearest  to  satisfying  the  criterion  for  maximum  moment  about  d 
(or  D). 

Then  by  taking  moments  about  J  we  obtain 


R  =  [  (16,364  x  75)  +  (284  x  105)  +  105  ]  =  254,260  lbs.  (about) 


DESIGN  OF  SIMPLE  RATLEOAD  BRIDGES 


445 


for  the  reaction  at  A.     Next,  taking  moments  about  d  of  the  forces  to  the 
left  and  dividing  by  the  height  of  the  truss  at  that  point  we  obtain 


S=  [(254,260  x  75)  -  (5,848  x  1,000)] 


1 


42.43 


=  311,600  Ibs.  (about) 


for  the  maximum  stress  in  chord  DE  due  to  the  -E40  loading  and  multiply- 
ing this  by  50/40  we  obtain  389,500,  say  390,000  Ibs.  for  the  maximum 
stress  due  to  the  E50  loading. 
For  the  impact  we  have 


390,000  =  222,856,  say,  223,000  Ibs. 


Now,  adding  together  the  above  live-load  stress  and  impact,  and  the 
dead-load  stress  given  in  Fig.  311,  we  obtain 

390,000  +  223,000  +  174,000  =  787,000  Ibs. 

for  the  total  maximum  stress  in  bottom  chord  DE. 

Top  Chord  cd.    Considering  the  forces  to  the  left  of  section  2-2  (Fig. 
323)  it  is  readily  seen  that  the  horizontal  component  of  the  maximum 


Fig.  323 

stress  in  top  chord  cd  is  equal  to  the  maximum  stress  in  bottom  chord  DE. 
So  by  multiplying  the  maximum  stress  in  DE  by  sec<£l  we  obtain 

390,000  x  1.016  =  396,240,  say,  396,000  Ibs. 

for  the  maximum  live-load  stress  in  top  chord  cd. 
For  the  impact  we  have 


396,000  =  226,000  Ibs.  (about). 


Now,  adding  together  the  above  live-load  stress  and  impact,  and  the 
dead-load  stress  given  in  Fig.  311,  we  obtain 

396,000  +  226,000  +  177,000  =  799,000  Ibs. 

for  the  total  maximum  stress  in  top  chord  cd. 

Bottom  Chord  EF.  The  stress  is  found  in  this  member  by  taking 
moments  about  panel  point  e  and  hence  the  stress  in  the  member  will 
be  a  maximum  when  the  moment  about  e  is  a  maximum. 

Placing  wheel  13  at  E,  as  shown  in  Fig.  324,  we  have  2,020  Ibs.  for 
average  unit  load  to  the  left  of  E  (or  e)  and  2,062  Ibs.  for  the  average 
unit  load  on  the  bridge.  This  position  of  the  load  comes  the  nearest  to 
satisfying  the  criterion  for  maximum  moment  about  e  (or  E). 


446  STRUCTURAL  ENGINEERING 

Then  taking  moments  about  J  we  obtain 

R  =  [16,364  +  (284  x  90)  x  90*]  M^  =  222,320  Ibs.  (about) 


for  the  reaction  at  A.     Next,  taking  moments  about  e  of  the  forces  to  the 
left  and  dividing  by  the  height  of  the  truss  at  that  point  we  obtain 

S=  [(222,320  x  100)  -  (7,668  x  1,000) ]j~= 325,743  Ibs. 

for  the  maximum  stress  in  bottom  chord  EF  due  to  the  £40  loading  and 

325,743  x  |9  =  407,178,  say,  407,000  Ibs. 

for  the  maximum  stress  due  to  the  -E50  loading. 
For  the  impact  we  have 

1=  (2253+°3OQ)  407,000 -232,571,  say  233,000  Ibs. 

Now  adding  together  the  above  live-load  stress  and  impact,  and  the 
dead-load  stress  given  in  Fig.  311  we  obtain 

407,000  +  233,000  + 184,000  =  824,000 

for  the  total  maximum  stress  in  bottom  chord  EF. 

Top  Chord  de.    Considering  the  forces  to  the  left  of  section  3-3  (Fig. 
324)   it  is  readily  seen  that  the  horizontal  component  of  the  maximum 


stress  in  top  chord  de  is  equal  to  the  maximum  stress  in  bottom  chord  EF. 
So  by  multiplying  the  maximum  stress  in  the  bottom  chord  EF,  which, 
as  found  above,  is  407,000  Ibs.,  by  sec<£2  we  obtain 

407,000  x  1.004  =  408,628,  say,  409,000  Ibs. 

for  the  maximum  live-load  stress  in  top  chord  de. 
For  the  impact  we  have 

/=  (aa53+0300)  409,000  =  234,000 Ibs.  (about). 

Now,  adding  together  the  above  live-load  stress  and  impact,  and  the 
dead-load  stress  given  in  Fig.  311,  we  obtain 

409,000  +  234,000  + 185,000  =  828,000  Ibs. 

for  the  total  maximum  stress  in  top  chord  de. 

Top  Chord  ef.     In  accordance  with  usual  practice  we  will  consider 
thfi  stress  in  top  chord  ef  to  be  equal  and  opposite  to  the  stress  in  bottom 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  447 

chord  EF.  This  would  be  absolutely  true  if  the  shear  in  panel  EF  were 
zero.  If  a  uniform  load  be  used  the  shear  would  be  zero  and  in  case 
wheel  loads  are  used  the  shear  is  slight  and  hence  the  usual  practice  is 
not  far  wrong  in  any  case. 

We  have  now  determined  all  of  the  stresses  in  one-half  of  the  truss 
and  as  the  structure  is  symmetrical  about  the  center  of  span  these  are  all 
the  stresses  necessary  for  designing  the  trusses.  We  can  next  write  the 
above  stresses  on  the  stress  sheet,  Fig.  328. 

197.  Designing  of  Members  in  Trusses.  —  As  the  intermediate 
posts  are  the  governing  members  (see  Art.  176)  we  will  consider  them 
first.  We  will  first  ascertain  as  to  whether  channels  can  be  used.  The 
longest  post  is  J74-L4  (Fig.  328),  which  has  a  length  of  about  536  ins. 
The  average  radius  of  gyration  of  15"  channels  is  about  5.4  (see  Table 
3).  Then  for  the  maximum  L/r  we  have 

536-5.4  =  99.2. 

The  maximum  allowed  for  L/r  for  main  members  by  the  specifications 
is  100.  So,  as  far  as  L/r  is  concerned,  15"  channels  can  be  used.  We 
will  next  ascertain  as  to  whether  the  area  of  the  channels  is  sufficient. 
Post  U2-L2  has  the  greatest  stress  and  hence  will  require  the  great- 
est area.  This  post  is  about  454  ins.  long.  Now  substituting  in  the 
column  formula,  using  the  average  radius,  we  have 


p  =  16,000  -  70         =  10,115  Ibs. 

o.4 

for  the  allowable  unit  stress.     Dividing  this  into  the  stress  in  the  column 
we  obtain 

221,000  -f  10,115  =  21.74  sq.  ins. 

for  the  required  area.     From  this  it  is  seen  that  channels  can  be  used. 

Post  U2-L2.  The  area  just  found  for  this  post  is  21.74n".  Use 
2—  [sl5"x40#=23.52n".  (The  15"x35#  channels  have  1.16n"  less 
area  than  required.) 

Post  U3-L3.  This  member  has  a  maximum  compressive  stress  of 
157,000  Ibs.  and  a  maximum  tensile  stress  of  80,000  Ibs.  Combining, 
according  to  the  specifications,  we  have 


157,000  +  197,000  Ibs. 

& 

for  the  maximum  compression  and 

8(y)00  +  80f°0  =  120,000  Ibs. 

2 

for  the  maximum  tension  to  be  considered  in  designing  the  member. 

The  length  of  the  member  is  about  509  ins.     So  substituting  in  the 
column  formula,  using  the  average  radius  (as  preliminary),  we  have 


p=  16,000  -70       -  =  9,400  Ibs.  (about) 


448  STRUCTURAL  ENGINEERING 

for  the  allowable  compressive  unit  stress.     Then  we  have 
197,000  4-9,40.0  =  20.96  sq.  ins. 

for  the  area  of  cross-section  required  for  compression,  and  for  the  area 
required  for  tension  we  have 

120,000  -^  16,000  =  7.5  sq.  ins. 

As  is  seen,  the  compression  governs.  Referring  to  Table  3,  it  is  seen 
that  2  —  [s  15"  x  35*  =20  58n"  have  the  nearest  to  the  required  area  for 
compression.  So  substituting  the  radius  of  these  channels  in  the  column 
formula  we  have 


16,000  -70        -  =  9,615  Ibs. 

o.oo 

for  the  required  unit  stress.     Dividing  this  into  the  maximum  compres- 
sion we  obtain 

197,000  -r  9,615  =  20.49  sq.  ins. 

for    the    required     area.       This     is     very    close    to     the     area    of    the 
2  —  [s!5"x35#   and  hence  these  channels  will  be  used. 

Post  U4-L4-  This  member  has  a  maximum  compressive  stress  of 
112,000  Ibs.  and  a  maximum  tensile  stress  of  24,000  Ibs.  Combining, 
we  have  124,000  Ibs.  compression  and  36,000  Ibs.  tension  to  be  consid- 
ered in  designing  the  member.  The  length  of  the  member  is  about  536  ins. 
Then,  using  the  radius  of  the  lightest  15"  channel  (as  the  stresses  are 
low),  we  have 


16,000-  70  ££-  =  9,324  Ibs. 
0.0,0 

for  the  allowable  unit  compressive  stress  on  the  member.  Dividing  this 
into  the  above  combined  stress  we  have 

124,000  -r  9,324  =  13.3  sq.  ins.  (about) 

for  the  required  area  for  compression.  For  the  area  required  for  tension 
we  have 

36,000  ^  16,000  =  2.25  sq.  ins. 

As  is  seen,  the  compression  governs.  2 — [s  15"  x  33  #  =  19.8n"  (which 
are  the  lightest  15"  channels)  will  be  used  for  this  member.  As  is 
seen,  there  is  an  excess  of  6.5n"  of  metal,  yet  we  cannot  do  better  if 
channels  are  used.  The  L/r  would  be  too  great  if  12"  channels  were 
used  instead  of  the  15"  channels. 

Hanger  U1-L1.     This  is  entirely  a  tension  member.     For  the  re- 
quired net  area  of  cross-section  we  have 

198,000  -r  16,000  =  13.37  sq.  ins. 
We  will  use  2—  [s  15"  x  33*  =  19.8-  (21/32  x4)  =17.18n"net. 


DESIGN  OF  SIMPLE  KAILKOAD  BEIDGES  449 

(In  this  case  the  metal  cut-out  of  the  flanges  is  considered.)  This  is  more 
section  than  needed  but  15"  channels  are  used  for  the  intermediate  posts 
and  for  the  sake  of  appearance  they  will  be  used  for  this  member. 

Diagonal  U1-L2.  For  the  required  net  area  we  have  386,000- 
16,000  =  24.12°"  net.  Use  2— bars  S"  x  1 J"  =  24.0°". 

Diagonal  U2-L3.  For  the  required  net  area  we  have  280,000  + 
16,000  =  17.5n"  net.  Use  2— bars  7"  x  1 J"  =  17.5n". 

Diagonal  U3-L4.  For  the  required  net  area  we  have  222,000  -r 
16,000  =  13.87°"  net.  Use  2— bars  7"  x  1  =  14n"  net. 

Diagonal  U4-L4.  For  the  required  net  area  we  have  171,000  -r 
16,000  =  10.68n"  net.  Use  4— Ls  5"  x  3J  x  |  =  12.20  -  1.50  =  10.7n"  net. 

Counter  U4~L3.  For  the  required  area  of  cross-section  we  have 
90.000-7- 16,000  =  5.62n".  Use  1— bar  2-f"  x2f"  =  5.64n"  (standard  bar). 

Bottom  Chords  LO-Ll  and  L1-L2.  For  the  required  net  area  we 
have  493,000  +  16,000  =  30.8n".  Use  the  following  section : 

2— pis.  20"  xf  "  =  25.0   -3.75  =21.25°"  net 

4— Ls  3 J"  x  3J"  x  Ty  =  11.48  -  1.75  =    9.73n"  net 


30.98n"  net 

Bottom  Chord  L2-L3.     For  the  required  area  of  cross-section  we 
have  687,000  -r  16,000  =  42.92°".     Use  the  following  section : 

2— bars  8"  x  If"    =  22.00°" 
2— bars  8"  x  lTy  =  21.00°" 


43.00°" 

Bottom   Chord  L3-L4-     For  the  required  area  of  cross-section  we 
have  787,000 ~  16,000  =  49.18°".     Use  the  following  section: 

2— bars  8"  x  1£"   =  24.00°" 
2— bars  8"  x  lTy  =  25.00°" 


49.00°" 

Bottom  Chord  L4-L4.  For  the  required  area  of  cross-section  we 
have  824,000  + 16,000  =  51.5°".  Use  4— bars  8"  x  If  =  52.0". 

In  designing  the  top  chords,  it  is  best  to  design  the  lightest  section 
first,  using  minimum  thicknesses  of  web  (if  sufficient)  so  that  the  area  of 
cross-section  can  be  increased  for  the  others  by  merely  increasing  the 
thickness  of  the  webs.  So  in  this  case  we  will  first  consider  top  chord 
C71-Z72,  as  it  has  the  least  stress  and  consequently  will  have  the  lightest 
section. 

Top  Chord  U1-U2.  The  first  thing  to  do  is  to  determine  the  gen- 
eral dimensions  of  the  section.  The  width  of  the  posts  will  be  12J" 
(see  Art.  176)  and  the  maximum  thickness  of  eye-bars  is  3"  (2— 8"x  if" 
bars  at  [71 )  and  allowing,  say,  2"  for  pin  plates  and  clearance  we  have 
12^  +  3  +  2  =  174-"  for  the  required  distance  between  the  webs.  Then  if 
the  webs  be  f"  thick  (each)  and  the  top  angles  be  4"x4"  we  obtain 
17J  +  li  +  8  =  26f,  say,  27"  for  the  width  of  the  cover  plate.  The  radius 
of  gyration  in  reference  to  the  y-y  axis  (see  Art.  176)  is  approximately 


450  STRUCTURAL  ENGINEERING 

equal  to  the  distance  from  the  center  of  the  cover  plate  out  to  the  outer 
face  of  the  web,  which  in  this  case  is  (17^  +  li)  -r  2  =  9.37.  Now,  in 
order  that  the  section  be  of  economic  depth,  that  is,  so  that  the  radius 
of  gyration  about  the  x-x  axis  is  the  same  as  about  the  y-y  axis,  the  webs 
should  have  a  depth  0.4  of  which  should  equal  the  above  radius.  So  for 
the  economic  depth  of  the  web  we  have 


So  we  will  make  the  webs  24"  deep.  According  to  the  specifications, 
the  thickness  of  the  cover  plate  should  not  be  less  than  ^  of  the  dis- 
tance between  the  rivet  lines.  The  distance  between  the  rivet  lines  in 
this  case  will  be  about  17J  +  1J  +  4^  =  23  J".  ^  of  this  distance  is 
about  -j96  of  an  inch.  So  we  will  make  the  cover  plate  -£%"  thick.  The 
thickness  of  the  webs,  according  to  the  specifications,  should  not  be  less 
than  ^  of  the  distance  between  the  flanges.  If  the  top  angles  be 
4"x4"  and  the  bottom  angles  6"x4"  (the  6"  leg  along  the  web),  we 
have  24-  (4  +  6)  =14"  for  the  distance  between  the  flanges.  Then  for 
the  minimum  thickness  of  the  webs  we  have  14  x  1/30  =  0.46,  which  is 
about  -£Q  of  an  inch. 

The  top  angles  should  be  of  minimum  thickness  and  the  bottom 
angles  should  be  of  maximum  thickness  in  order  that  the  center  of  gravity 
of  the  section  be  as  near  the  center  of  the  web  as  possible. 

The  length  of  the  member  (U1-U2)  is  about  312  ins.  Now  substi- 
tuting this  length  and  the  approximate  radius  given  above  in  the  column 
formula  we  obtain 


p  =  16,000  -  70        =  =  13,670  Ibs. 
9.o7 

for  the  approximate  allowable  unit  stress.     Dividing  this  into  the  maxi- 
mum stress  in  the  member  we  obtain 

713,000  -r  13,670  =  52.15  sq.  ins. 

for  the  approximate  required  area  of  cross-section. 

In  accordance  with  the  above  we  obtain  the  following  section: 


l_CoVer  pi.  27"xTV'  =  1 
2— web  pis.  24"xTy  =  21.00n" 
2— Ls4"x4"x£"  =  5.72n" 
2— Ls  6"  x  4"  x  Ty  =  10.62n// 

52.52n" 

•     This  section  is  really  the  minimum  and  is  very  near  the  approxi- 
mate section  found  above. 

Now  taking  moments  about  the  cover  plate,  in  the  manner  shown 
in  Art.  176  we  obtain  9.38"  for  the  distance  from  the  cover  plate  down 
to  the  horizontal  gravity  axis  of  the  section. 

The  next  thing  is  to  see  if  the  eye-bars  will  fit  into  the  top  chord. 
The  largest  pin,  which  will  be  at  Ul,  will  be  about  7"  in  diameter.  From 
the  table  of  eye-bars,  to  be  found  in  practically  all  structural  hand- 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  451 

books,  it  is  seen  that  the  8"  bars  require  17J"  head  for  a  7"  diameter 
pin.  Half  of  this  is  8|".  So  it  is  seen  that  the  eye-bars  fit  into  the 
chord  satisfactorily,  as  it  is  9.38"  from  the  gravity  axis  to  the  cover 
plate  and  hence  the  bars  will  not  interfere  with  the  plate. 

In  the  same  mariner  as  shown  in  Art.  176  the  radius  of  gyration 
about  the  horizontal  gravity  axis,  or  x-x  axis,  is  found  to  be  about  9.5  and 
9.3  about  the  y-y  axis.  Now  using  the  least  radius  we  obtain 

q-i  o 

16,000  -  70  -^-  =  13,652  Ibs. 

for  the  actual  allowable  unit  stress. 

Then  dividing  this  into  the  stress  we  obtain 

713,000^13,652  =  52.22 

for  the  required  area  of  cross-section  which  is  practically  equal  to  the 
above  assumed  section  and  hence  that  section  will  be  used. 

Top  Chord  U2-U3.  Dividing  the  maximum  stress  in  this  member 
by  the  allowable  unit  stress  found  for  C71-J72  (which  will  be  practically 
the  same  as  for  this  member)  we  obtain 

799,000  ^  13,652  =  58.53  sq.  ins. 

for  the  approximate  area  of  cross-section  required.  The  following  section 
has  about  this  area: 

1— cov.  pi.  2T"xTy  =  15.18n" 
2 — web  pis.  24"x1V'  =  27.00n" 
2— Ls4"x4"xf"  =  5.72n// 
2— Ls  6"  x  4"  x  A"  =  10.62n" 


58.52n" 

By  taking  moments  about  the  cover  plate  we  obtain  9.66"  for  the 
distance  from  the  cover  plate  down  to  the  horizontal  gravity,  or  x-x  axis. 
The  radius  of  gyrations  (found  as  previously  explained)  in  reference 
to  the  x-x  axis  is  9.3  and  the  radius  in  reference  to  the  y-y  axis  is  9.2. 

The  length  of  the  member  is  about  305  ins.  Now,  using  the  least 
radius  we  obtain 


16,000  -70  -        =  13,680  Ibs. 
y.,o 

for  the  allowable  unit  stress.  Dividing  this  into  the  maximum  stress 
in  the  member  we  obtain 

799,000  *  13,680  =  58.40  sq.  ins. 

for  the  required  area  of  cross-section.  This  is  practically  the  same 
as  the  approximate  section  found  above  and  hence  that  section  will  be 
used. 

Top  Chord  U3-U4.     Dividing  the  maximum  stress  in  the  member  by 
the  allowable  unit  stress  found  for  U2-U3  we  obtain 

828,000  -r  13,680  =  60.52  sq.  ins. 


452 


STRUCTURAL  ENGINEERING 


for  the  approximate  area  of  cross-section  required.     The  following  section 
has  about  this  area: 


1—  cov.    pi.    27"xTV'  =  1 
2_web  pis.  24"  xf"    =  30.00n" 

2—  Ls4"x4"xf"          =    5.72n" 
2_Ls  6"  x  4"  x  TV        =  10.62n// 

61.52n" 

The  allowable  unit  stress  for  this  member  is  practically  the  same 
as  for  U2-U3  and  the  above  section  is  as  near  the  required  section  as  we 
can  obtain  without  changing  the  thickness  of  the  bottom  angles  which 
would  be  an  objectionable  thing  to  do,  as  the  center  of  gravity  would  be 
shifted,  and  hence  the  above  section  will  be  used. 

Top  Chord  U4-U4*  As  the  allowable  unit  stress  for  this  member 
is  practically  the  same  as  for  [73-174  and  the  stress  is  practically  the  same 
the  same  section  will  be  used  as  for  [73-174. 

198.     Designing   of   the   Bottom   Lateral    System.  —  The   load 

specified  by  the  specifications  for  the  bottom  laterals  is  700  Ibs.  per  foot 
of  span.  This  load,  according  to  the  specifications,  is  to  be  considered 
as  a  live  load. 

For  the  panel  load  we  have  700x25  =  17,500  Ibs.,  and  for  determin- 
ing the  stresses  we  have  the  following: 

17,500  -=-9          =    1,944.44. 
Secw  1.78. 

1,944.44x1.78=   3,461.1. 


fe-z 


V  ^' 


X 


X 


Fig.  325 

Lateral  Ab.     For  the  maximum  tensile  stress  in  this  member   (see 
Fig.  325,  also  Art.  177)  we  have 

36  x  3,461.1  =  124,600,  say  125,000  Ibs. 
For  the  area  of  cross-section  required  we  have 

125,000  -f  16,000  =  7.81  sq.  ins. 

Use  2— Ls  6"  x  3J"  x  £"  =  8.0°"  net. 

Lateral  Be.     For  the  maximum  tensile  stress  in  this  member  we  have 

28  x  3,461.1  =  96,910,  say  97,000  Ibs. 
For  the  area  of  cross-section  required  we  have 

97,000  + 16,000  =  6.06  sq.  ins. 
Use  2— Ls  6"  x  3  J"  x  |"  =  6.09"  net. 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES 


453 


Lateral  Cd.  For  the  maximum  tensile  stress  in  this  member  we 
have 

21  x  3,461.1  =  72,683,  say  73,000  Ibs. 

For  the  area  of  cross-section  required  we  have 

73,000  -f  16,000  =  4.56  sq.  ins. 

Use  2—  Ls  34"  x  34"  x  //'  =  4.87n"  net. 

Lateral  De.     For  the  maximum  tensile  stress  in  this  member  we  have 

15  x  3,461.1  =  51,916,  say  52,000  Ibs. 
For  the  area  of  cross-section  required  we  have 

52,000  +  16,000  =  3.25  sq.  ins. 

Use  2—  Ls  34"  x  34"  x  f  "  =  4.21n"  net. 

Lateral  Ef,    For  the  maximum  tensile  stress  in  this  member  we  have 

10  x  3,461.1  =  34,611,  say  35,000  Ibs. 
For  the  area  of  cross-section  required  we  have 

35,000  -=-  16,000  =  2.18  sq.  ins. 

Use  2—  Ls  34"  x  3^"  x  f"  =  4.21n"  net. 

This  completes  the  designing  of  the  bottom  laterals  as  the  structure 
is  symmetrical  about  the  center  of  span. 

The  stringer  bracing  is  designed  as  previously  explained  for  deck 
plate  girder  bridges  and  the  transverse  struts  connecting  to  the  stringers 
iit  the  points  of  intersection  of  the  laterals  are  designed  as  explained  in 
Art.  177  for  the  150-ft.  riveted  bridge. 

199.  Designing  of  Top  Lateral  System.—  The  load  specified  by 
the  specifications  for  the  top  laterals  is  200  Ibs.  per  foot  of  span.  This 
load,  according  to  the  specifications,  is  to  be  considered  as  a  live  load. 

For  the  panel  load  we  have  P  -  25  x  200  =  5,000  Ibs.,  and  for  deter- 
mining the  stresses  we  have  the  following: 

5,000  -=-9  =  555.55. 


Seco>3  =  1.78. 
Secw    =1.78. 


®       ©        ©       ®       <D        0 

Fig.  326 

Lateral  Be.     For  maximum  tensile  stress  in  this  member  (see  Fig. 
326)  we  have  (see  Art.  178) 

-  secwl  x  28  =  555.55  x  1.82  x  28  =  28,310,  say,  28,000  Ibs. 


454  STRUCTURAL  ENGINEERING 

For  the  area  of  cross-section  required  we  have 

28,000  ~  16,000  =  1.75  sq.  ins. 

Use  2—  Ls  3J"  x  3  J"  x  f"  =  4.22*"  net.     These  are  about  the  smallest  an- 
gles used  in  railroad  work  and  hence  are  used  in  this  case. 

Lateral  Cd.     For  the  maximum  tensile  stress  in  this  member  we  have 

z> 

£-  seco>2  x  21  =  555.55  x  1.79  x  21  =  20,883,  say  21,000  Ibs. 

u 

For  the  area  of  cross-section  required  we  have 

21,000  -f  16,000  =  1.31  sq.  ins. 


Use  2—  Ls3J"x3J"xf"  =  4.22n"net—  same  as  for  Be. 

Lateral  De.    For  the  maximum  tensile  stress  in  this  member  we  have 

£  secw3  x  15  =  555.55  x  1.78  x  15  =  14,883,  say  15,000  Ibs. 
y 

Use  the  same  section  as  used  for.  Be—  2—  Ls  3J"  x  3±"  x  f". 

Lateral  Ef.    For  the  maximum  tensile  stress  in  this  member  we  have 

£  secco  x  10  =  555.55  x  1.78  x  10  =  9,888,  say  10,000  Ibs. 

y 

Use  the  same  section  as  used  for  Be—  2—  Ls  3  \"  x  3  J"  x  f  ". 

Strut  Cc.     For  the  maximum  compression  in  this  member  we  have 
(see  Art.  178) 

21  x  555.55  +  (  ~^-  \  =  14,166,  say,  14,000  Ibs. 
\      X      / 

The  length  of  member  (c.c.  of  chords)  is  17  ft.,  or  204  ins.  If  the  strut 
be  composed  of  4  —  Ls  3-J-"  x  3"  x  §"  the  least  radius  of  gyration  will  be 
about  1.7.  Then  L/r  =  204  -f  1.7  =  120,  which  is  just  the  maximum  limit 
permitted  by  the  specifications  and  hence  the  above  angles  will  be  used. 
It  is  seen  that  the  stress  in  this  strut  is  so  low  that  it  does  not  really 
influence  the  design,  and,  as  the  stress  in  the  other  struts  —  Dd,  Ee,  etc.  — 
are  less  yet,  there  is  really  no  need  of  computing  it.  We  will  simply 
make  each  of  the  other  struts  of  4—  Ls  3£"  x  3"  x  f  "  and  let  it  go  at  that. 
200.  Design  of  Portal.  —  Assuming  the  ends  of  the  end  posts  fixed 
and  using  the  same  symbols  as  used  in  Art.  179  (see  Fig.  271)  we  have 
in  this  case 

P  =  5,000  Ibs., 

R  =  5,000  x  28  =  15,555  Ibs., 

t/ 

H  =  10,277  Ibs., 
±V=  (20,555  x  24.16)  -s-  17  =  29,212,  say  29,000  Ibs. 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES 
Taking  moments  about  p  (Fig.  327)  we  have 


455 


and  substituting  the  numerical  values  given  above  and  reducing  we  obtain 
S  =  18,055  +  (10,277  x  15,665)  -=-  8.5  =  36,994,  say  37,000  Ibs. 

for  the  compressive  stress  in  the  part  Bo  of  the  portal. 

In  a  similar  manner  taking  moments  about  t  we  obtain 

SI  =  2,500  +  18,939  =  21,439,  say  22,000  Ibs. 

for  the  tensile  stress  in  the  part  ob  of  the  portal. 

For  the  stress  in  the  parts  po  and  ot  of  the  portal  we  have 

S2  =  V  secc/.  =  29,000  x  1.42  =  41,180,  say  41,000  Ibs. 

This  is  compression  in  one  and  tension  in  the  other.  With  the  applied 
forces  acting  as  indicated  in  Fig.  327,  op  would  be  in  tension  and  ot  in 
compression. 

The  designing  of  the  sections  of  the  members  of  the  portal  in  this 
case  is  mostly  a  matter  of  obtaining  rigidity  as  the  stresses  are  relatively 
low.  For  each  of  the  members 
Bbf  op  and  ot  we  will  use 
2—  Ls  4"  x  4"  x  f  "  and  for  eac?" 
of  the  secondary  members,  the 
ones  shown  dotted  in  Fig.  327, 
we  will  use  2—  Ls  3|"x3J"xf". 
These  sections  are  obtained  in 
the  manner  shown  in  Art.  179. 
201.  Design  of  End  Post. 
—  As  stated  in  Art.  180,  colli- 
sion struts  are  desirable,  but 
they  are  sometimes  omitted  and 
for  the  sake  of  variety  we  will 
omit  them  in  this  bridge.  The 
cross-section  of  the  end  posts 
will  be  the  same  in  form  as 
that  of  the  top  chords.  The  Fig.  327 

length   of   the   member   as   re- 

gards the  x-x  axis  (see  Fig.  265)  will  be  taken  as  the  full  length, 
which  is  478  ins.,  and  as  regards  the  y-y  axis  the  length  will  be  taken  as 
the  distance  from  the  lower  end  of  the  end  post  to  the  portal  (distance 
pu,  Fig.  327).  This  distance  is  about  376  ins. 

Using  the  radius  in  reference  to  the  x-x  axis  found  for  the  chord 
section  £72-173  as  preliminary  we  obtain 


=  16,000  - 


4-7fi 


=  12,403 


for  the  approximate  allowable  unit  stress,  provided  there  were  no  cross 
bending  on  the  member.      Dividing  this   unit   stress   into  the   maximum 


456  STRUCTURAL  ENGINEERING 

stress  in  the  member  we  obtain  782,000 -r  12,403  =  63.04°".  Now  as  the 
member  is  subjected  to  cross  bending,  due  to  wind,  the  section  should 
very  likely  be  larger  than  this.  Let  us  assume  the  following  section : 

1— cov.  pi.  27"xTV  =  15.1Sn" 
2— web  pis.  24"  xlJ"  =  33.00°" 
2— Ls4"x4"xi"  =  7.50n" 
2— Ls  6"  x  4"  x  f"  =  11.72°" 


67.40n" 

Taking  moments  about  the  cover  plate  of  this  section  it  is  found 
that  the  distance  from  the  cover  plate  to  the  gravity  axis  x-x  is  10.14  ins. 

The  radius  of  gyration  in  reference  to  the  x-x  axis  is  9.22.  The 
moment  of  inertia  in  reference  to  the  y-y  axis  is  5,742  and  the  radius  in 
reference  to  the  same  axis  is  9.23. 

Then  for  the  allowable  unit  stress,  provided  there  were  no  cross  bend- 
ing we  would  have 


16,000-  70       -  =  12,371  Ibs. 

&•&& 

But  in  case  of  direct  and  bending  stresses  combined  this  can  be  increased 
25  per  cent  (see  specifications).  So  we  have 

12,371x1.25  =  15,464  Ibs. 

for  the  allowable  unit  stress. 

The  maximum  bending  moment  on  the  end  post  due  to  wind,  consid- 
ering the  posts  fixed  at  the  ends,  will  occur  at  the  bottom  of  the  portal; 
that  is,  at  points  p  and  t  (Fig.  327).  For  this  moment  we  have 

M  =  10,277  x  15.665  x  12  =  1,931,870  inch-lbs. 

(see  Fig.  327). 

Then  for  the  maximum  bending  stress  we  have 

1,931,870 
/=  '   —  x  13.  5  =  4,542  Ibs.  persq.  in. 

O  j  I  *X  /V 

For  the  actual  direct  unit  stress  we  have 

782,000  -=-67.4  =  11,602  Ibs. 
Now  adding  the  bending  stress  to  the  direct  stress  we  have 

4,542  +  11,602  =  16,144  Ibs. 

for  the  actual  unit  stress  on  the  member  due  to  direct  stress  and  cross 
bending  combined.  As  this  stress  is  very  near  the  allowed,  being  only 
680  Ibs.  more  (16,144-15,464  =  680),  the  above  assumed  section  will 
be  used. 

202.  Maximum  Reaction  on  Shoe.  —  The  dead-load  reaction  on  the 
shoe  is  the  same  as  for  the  truss  except  for  the  half  panel  load  at  the  end. 


DESIGN  OF  SIMPLE  EAILKOAD  BEIDGES  457 

The  panel  load  of  dead  load  is  33,000  Ibs.  (see  Art.  195).  Then  for 
the  dead-load  reaction  on  the  shoe  we  have 

4J  x  33,000  =  148,500  Ibs. 

Placing  wheel  2  at  one  end  of  the  span,  as  shown  in  Fig.  278  (Art. 
182)  for  the  150-ft.  span,  and  taking  moments  about  the  other  end  we 
obtain  365,000  Ibs.  for  the  maximum  live-load  reaction  on  the  shoe.  Multi- 
plying this  by  300  ~  (225  +  300)  we  obtain  208,500  Ibs.  for  the  impact. 
Then  adding  together  the  above  dead-  and  live-load  reactions  and  impact 
we  obtain 

148,500  +  365,000  +  208,500  =  722,000  Ibs. 

for  the  total  maximum  reaction  on  the  shoe. 

The  stress  sheet,  Fig.  328,  can  now  be  completed  and  then  the  work 
of  designing  the  details  can  proceed. 

203.  Details. —  The  general  details  of  the  entire  span  are  shown  in 
Figs.  329,  330  and  331.  These  details  are  drawn  to  a  f"  scale  on  f" 
layout;  that  is,  the  outline  of  the  truss  is  drawn  to  a  f"  scale  and  the 
details  are  drawn  on  this  layout  to  a  f "  scale. 

Joint  LO.  The  detail  of  this  joint  is  shown  in  Fig.  329.  In 
drawing  this  joint,  first  the  joint  line  should  be  located,  which  is  done 
by  bisecting  the  angle  between  the  end  post  and  shoe.  Then  the  distance 
from  the  cover  plate  of  the  end  post  to  the  center  line  of  pin  should  be 
determined  as  explained  in  Art.  184.  Then  the  outline  of  the  lower 
end  of  the  end  post  can  be  drawn  and  next  the  outline  of  the  shoe  can 
be  sketched.  The  shoe  should  be  deep  enough  to  resist  the  cross  bend- 
ing on  it,  as  explained  in  Art.  142,  and  it  should  be  long  enough  to  cover 
the  rollers.  The  general  dimensions  of  the  shoe  and  number  of  rollers 
are  first  assumed  and  then  modified,  if  necessary,  to  agree  with  the 
calculated  requirements. 

By  drawing  the  end  view  and  part  cross-section  of  the  joint,  as 
shown,  to  the  left  of  I/O,  about  the  desired  length  of  the  rollers  is  obtained. 
The  size  of  the  rails  in  the  pedestal  is  selected,  the  rails  are  spaced  and 
then  a  suitable  diameter  of  roller  can  be  determined.  In  this  case,  as  is 
seen,  70-lb.  (per  yd.)  rails  are  used.  Allowing  for  one-eighth  of  an  inch 
to  be  planed  off  the  top  of  these  rails,  the  width  of  bearing  of  each 
against  a  roller  is  2J"  (see  Carnegie  and  the  Illinois  Steel  Company 
catalogue) . 

Resolving  the  stress  in  the  end  post  vertically  (which  can  be  done 
very  rapidly  by  graphics)  we  obtain  612,000  Ibs.  (about).  One-half  of 
this,  which  is  306,000  Ibs.,  should  be  supported  by  the  rails  to  the  left 
of  the  center  line  of  the  end  post.  Assuming  7"  rollers  (as  shown),  the 
allowable  bearing  per  inch  of  length  of  roller  is  7  x  600  =  4,200  Ibs.  Then 
as  there  are  five  rails  (not  considering  the  90-lb.  guide  rail  at  the  center 
line)  to  the  left  of  the  center  line  of  the  end  post  and  there  being  seven 
rollers,  we  have 

(5  x  2i)  7  x  4,200  =  330,700  Ibs. 

for  the  allowable  pressure  on  the  part  of  the  rollers  to  the  left  of  the 
center  line  of  the  end  post,  which,  as  is  seen,  is  about  24,000  Ibs.  more 
than  necessary.  The  first  five  rails  to  the  right  of  the  center  line  of  the 


458  STRUCTURAL  ENGINEERING 

end  post  will  support  the  other  half  of  the  vertical  component  of  the  end 
post  and  the  other  two  rails,  directly  under  the  end  floor  beam,  should 
be  about  sufficient  to  support  the  maximum  reaction  of  the  beam.  This 
reaction  (see  Fig.  328)  is  143,500  Ibs.  For  the  allowable  pressure  on 
the  two  rails  we  have 

(2  x  2  J)  7  x  4,200  =  132,300  Ibs., 

which  is  11,200  Ibs.  less  than  required,  but  as  the  maximum  stress  in 
the  end  post  and  the  maximum  reaction  on  the  end  floor  beam  do  not  occur 
at  the  same  time  (and  the  other  rails  are  understressed),  the  design  of 
the  roller  pedestal,  as  shown,  is  satisfactory. 

The  shoe  proper  must  be  made  to  conform  with  the  details  of  the 
end  post.  Assuming  a  1"  pin,  we  obtain 

782,000  -=-  (24,000  x  7)  =  4.65  ins. 

for  the  required  thickness  of  pin  bearing  on  the  end  post,  or  2.32"  on 
each  side.  We  have  on  each  side  of  the  post,  as  shown,  a  •&",  \"  and  £ " 
plate  and  an  \\"  web,  making  in  all  2§"  (2.375)  of  bearing,  which  is  the 
thickness  required. 

For  the  required  thickness  of  bearing  on  the  shoe  we  have 

612,000  -=-  (24,000  x  7)  =  3.64  ins., 

or  1.82"  at  each  side.  We  have  at  each  side,  as  shown,  3 — f"  plates, 
making  1J"  (1.875)  bearing  which  is  about  the  required  bearing.  It 
will  be  seen  that  in  addition  to  this  bearing  there  is  a  £ "  filler  and  a 
A"  Jaw  plate,  but  as  these  extend  only  a  short  distance  below  the  pin, 
they  will  not  be  considered  for  bearing. 

The  bottom  edges  of  the  vertical  bearing  plates  of  the  shoe  are 
planed  so  that  they  bear  firmly  against  the  sole  plate  and  hence  the  rivets 
passing  through  the  plates  simply  hold  them  together.  However,  there 
should  be  as  many  rivets  in  the  vertical  legs  of  the  angles  connecting 
these  plates  to  the  sole  plate  as  is  possible  to  put  in  so  that  the  pressure 
will  be  well  distributed  over  the  sole  plate  by  the  outstanding  legs  of 
the  angles. 

The  pin  plates  on  the  lower  end  of  the  end  post  should  be  arranged 
so  that  they  firmly  grip  the  post.  For  the  allowable  bearing  on  the  •£%" 
inside  plate  against  the  7"  pin  we  have 

A  x  7  x  24,000  =  94,500  Ibs. 

Then  the  number  of  J"  shop  rivets  required  in  single  shear  to  hold  this 
plate  is 

94,500 -r  7,200  =  13. 

As  is  seen,  there  are  21.  For  the  allowable  bearing  on  the  J"  outside 
plate,  we  have 

£x  7x24,000  =  84,000  Ibs. 

Then  for  the  number  of  J"  shop  rivets  required  in  single  shear  to  hold 
this  plate  we  have 

84,000-7,200  =  12. 


459 


L"" 


Fig.  329 

460 


22Sfl  Thro.  S.  Ypin  'ConnectedSpan 
General Dran'ng 


'(//a.  -r 
Spedrtcofions,  A  ft£.  Asm, 


Fig.  831 


462 


DESIGN  OF  SIMPLE  RAILKOAD  BEIDGES  463 

This  plate  should  have  a  sufficient  number  of  rivets  connecting  it  directly 
to  the  top  and  bottom  angles  of  the  end  post  to  transmit  the  total  pin 
pressure  on  it  to  these  angles.  There  are  14  passing  through  it  and  the 
angles,,  so  the  riveting  is  satisfactory,  although  counting  the  other  rivets 
passing  through  the  plate  there  is  an  excess  of  15  rivets.  For  the  allow- 
able bearing  on  the  f  "  filler  we  have 

f  x  7x24,000  =  105,000  Ibs. 

Then  for  the  number  of  J"  shop  rivets  required  in  single  shear  to  hold 
the  filler  we  have 

105,000-^7,200  =  15. 

As  is  seen,  there  are  23  passing  through  the  filler,  which  is  an  excess  of 
8  rivets.  The  ^"  outside  plate  and  the  f"  filler  acting  together  tend  to 
shear  the  rivets  off  against  the  web  and  angles  of  the  end  post.  The 
combined  allowable  bearing  of  these  two  plates  against  the  7"  pin  is 

84,000  +  105,000  =  189,000  Ibs. 

Then  for  the  number  of  J"  shop  rivets  required  to  hold  these  two  plates 
when  considered  as  acting  together  we  have 

189,000  -r  7,200  =  27. 

As  is  seen,  there  are  37  —  an  excess  of  10  rivets. 

Although  from  the  above  there  appears  to  be  an  excess  of  rivets  in 
these  pin  plates,  nevertheless,  the  rivets  are  spaced  about  as  sparingly 
as  possible,  while  yet  obtaining  good  details,  so  the  riveting  is  satisfactory. 

According  to  the  specifications,  the  net  area  of  cross-section  through 
the  pin  hole  of  the  bottom  chord  (at  Z/0)  should  be  25  per  cent  more 
than  the  net  section  of  the  member.  The  net  section  of  the  member  (see 
Fig.  328)  is  30.98n".  Then  the  area  of  section  through  the  pin  hole 
should  be  38.72n". 

As  is  seen,  the  TV'  plates  have  a  net  area  through  the  pin  hole  of 
5.25n",  the  i"  plates  have  13.00n",  and  the  web  has  16.25n",  making  a 
total  of  34.51n"  for  the  plates.  In  addition,  each  of  the  angles  can  be 
considered  to  the  extent  of  four  rivets  in  shear.  Each  angle  then  is 
equivalent  to  (4  x  7,200)  -^-16,000  =  1.8n",  or  7.2n"  for  the  four  angles. 
Adding  this  to  the  net  area  of  the  plates  we  obtain  34.51  +  7.2  =  41.71°", 
which  is  2.99n"  more  than  required. 

According  to  the  specifications  (A.  R.  E.  Ass'n),  the  net  area  along 
the  center  line  between  the  pin  and  the  end  of  the  member  must  be  equal 
to  the  net  area  of  cross-section  of  the  member.  As  is  seen,  we  have 


for  the  net  area  along  the  center  line  between  the  pin  hole  and  the  end 
of  the  member.  As  this  is  practically  equal  to  the  net  area  of  cross-sec- 
tion of  the  member,  which  is  30.98n",  the  detail  of  the  end  of  the  bottom 
chord,  as  shown,  is  satisfactory  as  far  as  sections  are  concerned. 

Considering  the  net  area  of  cross-section  through  the  pin  hole,  the 
strength  of  the  Ty  outside  plate  in  tension  is   Gx^  x  16,000  =  42,000 


464 


STRUCTURAL  ENGINEERING 


Ibs.  Then  for  the  number  of  J"  shop  rivets  required  to  the  right  of  the 
pin  hole  to  transmit  this  stress  in  single  shear  we  have 

42,000 -=-7,200  =  5.8,  say  6. 

As  is  seen,  6  rivets  are  used. 

The  strength  of  the  \"  inside  plate  in  tension  is  13  x-|x  16,000 
=  104,000  Ibs.  Then  for  the  number  of  J"  shop  rivets  required  to  the 
right  of  the  pin  hole  to  transmit  this  stress  in  single  shear,  we  have 

104,000*7,200  =  15. 

As  is  seen,  16  rivets  are  used.  From  the  above,  it  is  seen  that  the  detail 
of  the  end  of  the  bottom  chord  at  LO  is  correct  both  as  to  section  and 
riveting  and  hence  the  detail  is  satisfactory. 

Resolving  the  forces  in  the 
end  post  horizontally  and  verti- 
cally, we  have  the  loading  on  the 
pin  at  LO  shown  at  (a),  Fig.  332. 
Considering  first  the  bending  mo- 
ment on  the  pin  due  to  the  hori- 
zontal forces,  shown  at  (b),  the 
bending  moment  at  d  is  zero  and 
at  e  it  is  equal  to  0  +  246,500  x  2  = 
493,000"*,  which  is  the  maximum 
horizontal  moment,  the  forces 
being  symmetrical  in  reference  to 
the  center  line  cc. 

The  maximum  bending  mo- 
ment on  the  pin  due  to  the  vertical 
forces  shown  at  (c),  as  is  readily 


(Hoc) 


Fi     332  seen,  occurs  at  both  g  and  h  and  is 

equal  to   306,000  x  J  =  229,500"*. 
Then  for  the  resultant  maximum  bending  moment  we  have 


M  =    1493,000  *  +  229,500*=  543,000  inch  Ibs. 

Applying  Formula  (1),  Art.  83,  taking  /  as  25,000  Ibs.,  we  find 
that  a  6^"  pin  is  required  for  bending.  For  the  maximum  shear  on  this 
pin  we  have 


306,000 


(?'= 


10,385  Ibs.  per  sq.  in. 


So  the  6J"  is  large  enough  as  far  as  shear  is  concerned,  as  12,000  Ibs. 
per  square  inch  is  allowed.  The  bearing  as  computed  above  is  for  the  7" 
pin  and  to  use  a  6J"  pin  would  necessitate  increasing  the  bearing  by 
about  one-fourth,  which  would  increase  the  weight  of  metal  twice  as  much 
as  would  be  saved  on  the  pin  by  reducing  it  to  6^"  diameter.  So  there 
is  really  no  economy  in  using  the  smaller  pin  and  therefore  the  7"  pin 
will  be  used. 

The  other  details  at  LO  are  considered  to  be  self-explanatory,  espe- 
cially after  Arts.  184  and  186  are  carefully  read. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  465 

Joint  Ul.  In  drawing  this  joint  the  first  thing  to  do  is  to  locate 
the  joint  line  by  bisecting  the  angle  between  the  end  post  and  top  chord 
and  then  the  outlines  of  the  members  can  be  drawn  and  the  portal  located 
as  explained  in  the  case  given  in  Art.  184;  and  next  the  required  bearing 
on  the  pin  can  be  calculated. 

We  will  assume  a  1"  pin — the  size  used  at  LO;  then  for  the  required 
thickness  of  pin  bearing  on  the  end  post  we  have 

782,000  +  (24,000  x  7)  =  4.65  ins., 

which  is  about  4JJ".  As  is  seen,  we  have  on  each  side  of  the  post  a 
Ty  and  a  f"  plate  and  a  f"  filler.  This  with  the  H"  web  makes  2f" 
bearing  on  a  side  or  4^"  for  the  member,  which  is  practically  the  required 
bearing. 

For  the  required  thickness  of  pin  bearing  on  the  top  chord  we  have 

713,000  -r  168,000  =  4.24  ins. 

As  is  seen,  we  have  on  each  side  of  the  chord  1 — \"  and  2  —  -jV  plates 
and  1—  -f-g"  filler.  This,  with  the  ^V'  web,  makes  2f "  bearing  on  a  side 
or  4f "  for  the  member.  This  is  about  V  more  than  required,  but  it  is 
necessary  to  use  this  thickness  in  order  to  pack  the  joint. 

At  least  one  pin  plate  in  each  case  should  extend  at  least  6  ins. 
beyond  the  end  of  the  tie  plate. 

The  number  of  rivets  required  to  connect  the  pin  plates  to  the  end 
post  is  determined  in  the  same  manner  as  shown  above  for  joint  LO. 
For  the  allowable  pin  pressure  on  the  TV'  plate  we  have 

7  x&x  24,000  =  73,500  Ibs. 

Then  for  the  number  of  f  "  shop  rivets  required  to  transmit  this  in  single 
shear,  we  have 

73,500 -r7,200  =  ll  (about). 

It  is  obvious  that  most  of  the  pin  pressure  on  this  TV  outside  plate  will 
be  first  transmitted  to  the  f "  plate,  then  to  the  f "  filler,  and  on  to  the 
web  and  angles  of  the  end  post  instead  of  being  transmitted  directly  to 
the  web  and  angles  by  the  rivets,  as  the  rivets  are  quite  long  and  conse- 
quently will  bend  to  some  extent.  Let  us  assume  that  the  total  pin 
pressure  on  the  -f$"  plate  is  transmitted  to  the  f "  plate.  Then  for  the 
total  pressure  on  the  f "  plate  we  have 

73,500  +  7  x  f  x  24,000  =  178,500  Ibs. 

For  the  number  of  J"  shop  rivets  required  to  transmit  this  force,  we 
have 

178,500 -7,200  =  25  (about). 

There  should  be  about  enough  rivets  connecting  this  f "  plate  to  the  angles 
of  the  end  post  to  transmit  this  178,500  Ibs.  directly  to  the  angles  in 
order  to  distribute  the  pin  pressure  well  .over  the  cross-section  of  the 
member.  As  is  seen,  there  are  22  rivets,  which  is  about  the  correct  num- 
ber. For  the  allowable  pin  pressure  (bearing)  on  the  f "  filler  we  h&ve 

7  xf  x  24,000  =  105,000  Ibs. 


_/f.G6  STRUCTURAL  ENGINEERING 

Then  for  the  number  of  |"  shop  rivets  required  to  transmit  this  pressure 
in  single  shear,  we  have 

105,000 -5-7,200  =  14.5,  say  15. 

As  is  seen,  there  are  16  rivets  between  the  lower  end  of  the  filler  and 
the  -jV'  plate,  which  we  can  consider  as  holding  the  filler. 

As  is  seen  from  the  above,  if  the  rivets  just  below  the  pin  be  consid- 
ered only  to  take  the  pressure  on  the  TV'  outside  plate,  the  riveting  of 
the  pin  plates  to  the  end  post  at  £71  is  about  correct. 

The  allowable  pin  pressure  on  the  TV'  inside  pin  plate  of  the  top 
chord  is  7  x  T7Fx  24,000  =  73,500  Ibs.  It  requires  about  10  —  J"  shop 
rivets  to  transmit  this  in  singular  shear.  As  is  evident,  practically  all 
of  the  73,500  Ibs.  will  be  transmitted  first  to  the  £"  inside  pin  plate 
and  from  there  on  to  the  web,  and  hence  there  should  be  a  sufficient 
number  of  rivets  connecting  the  -J"  plate  to  the  web  to  transmit  the 
combined  pin  pressure  exerted  on  both  the  •£$"  and  -J"  plate. 

For  the  allowable  combined  pin  pressure  of  the  two,  we  have 

73,500  +  (7  x  \  x  24,000)  =  157,500  Ibs. 

It  requires  about  22 — J"  shop  rivets  to  transmit  this  in  single  shear.  As 
is  seen,  there  are  38  passing  through  the  -J"  inside  plate.  But  10  of 
these  should  be  considered  as  holding  the  TV'  inside  plate,  thus  really 
leaving  28,  which  is  6  more  than  required. 

The  allowable  pin  pressure  on  the  TV'  outside  pin  plate  is  73,500 
Ibs.  To  transmit  this  in  single  shear  requires  about  10 —  J"  shop  rivets. 
As  this  plate  connects  directly  to  the  angles  of  the  chord,  there  should 
be  enough  rivets  connecting  it  to  the  angles  to  transmit  the  entire  73,500 
Ibs.  As  is  seen,  there  are  14  rivets  connecting  the  plate  to  the  angles. 
So  the  riveting  is  satisfactory  as  far  as  the  TV'  outside  plate  is  con- 
cerned. The  allowable  pin  pressure  on  the  -f^"  filler  is 

7  x  T%  x  24,000  =  94,500  Ibs. 

To  transmit  this  in  single  shear  requires  about  13  —  J"  shop  rivets. 
As  is  seen  there  are  26 — twice  as  many  as  needed.  As  the  web  of  the 
top  chord  is  only  TV'  thick,  it  is  a  question  as  to  whether  the  rivets 
connecting  the  pin  plates  to  the  member  are  not  over-stressed  in  bearing 
on  the  web. 

The  \"  inside  plate  has  a  total  pressure,  as  given  above,  of  157,500 
Ibs.  There  are  38-10  (the  10  being  considered  to  hold  only  the  TV 
inside  plate)  to  transmit  this,  which  stresses  each  rivet  157,500 -r  28 
=  5,625  Ibs.  The  pin  pressure  on  the  TV'  outside  filler  is  94,500,  as 
given  above,  and  this  stresses  each  rivet  94,500 -r  26  =  3,634  Ibs.  Adding 
these  values  we  have  5,625  +  3,634  =  9,259  Ibs.  for  the  maximum  rivet 
bearing  on  the  TV'  web-  For  the  maximum  allowable  bearing  of  a  J" 
shop  rivet  on  •£%"  metal,  we  have 

TVxJx24,000  =  9,180  Ibs., 

which  is  about  the  bearing  exerted.  So,  taking  all  in  all,  the  riveting  of 
the  pin  plates  of  the  top  chord  at  £71  is  about  correct. 

The  calculations  for  the  details  at  the  end  of  the  hanger  at  £71  are 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES 


467 


made  in  the  manner  as  shown  above  for  the  end  of  the  bottom  chord 
at  LO. 

The  net  section  of  the  member  is  17,18n".  Then  the  net  section 
through  the  pin  hole  should  be  17.18  (1  +  0.25)  =21.57°". 

The  net  area  of  the  two  channels  through  the  pin  hole  is 
19.9  -  (7  x  0.4  x  2)  =  14.3°".  The  net  area"  of  the  2  —  J"  plates  through 
the  pin  hole  is  (12.5  x  \  x  2)  -  (7  xj  x  2)  =5.5n//  and  of  the  f"  plates 
it  is  6.87°",  making  in  all  14.3  +  5.5  +  6.87  =  26.67°",  which  is  about 
5°"  more  than  necessary.  , 

The  net  section  along  the  center  line  between  the  pin  hole  and  the 
end  of  the  hanger  should  be  equal  to  the  net  section  of  the  member,  or 
17.18°",  according  to  the  specifications.  We  have  (3.05x9.25) 
-  (3.05x3.5)  =17.54°",  which  is  about  the  section  required,  and  as 
the  distance  shown  as  9^"  can  not  be  increased,  on  account  of  clearance, 
the  detail  of  the  end  of  the  hanger  at  Ul  is  satisfactory  as  far  as  the 


Fig.   333 


section  is  concerned.  The  strength  of  the  J"  plate  in  tension  is 
16,000  x  5. 5  =  88,000  Ibs.  To  transmit  this  in  single  shear  requires  about 
12  —  J"  shop  rivets,  or  6  on  a  side,  which  should  be  below  the  pin.  As 
seen,  there  are  8.  The  strength  of  the  f "  plate  is  16,000  x  6.87  =  109,900 
Ibs.  This  requires  about  15  —  J"  shop  rivets,  or  7.5  on  a  side.  As 
seen,  there  are  8  passing  through  this  plate  outside  of  the  \"  plate. 
So,  taking  all  in  all,  the  detail  of  the  end  of  the  hanger,  as  shown  at  Ul, 
is  satisfactory. 

In  determining  the  bending  moment  on  the  pin  at  LO,  there  is  no 
question  concerning  the  live-load  stress  in  the  members  as  the  maximum 
in  the  end  post  and  bottom  chord  occurs  simultaneously,  but  at  Ul  the 
case  is  different,  for  at  that  joint  the  maximum  stress  in  the  members 
does  not  occur  simultaneously,  and,  consequently,  it  is  necessary  to  ascer- 
tain the  position  of  the  live  load  for  maximum  bending  moment  on  the 
pin  at  that  joint.  This  can  be  done  only  by  trial,  as  the  bending  on  the 


468  STRUCTURAL  ENGINEERING 

pin  due  to  any  member  not  only  depends  upon  the  stress  in  the  member, 
but  upon  its  lever  arm  as  well. 

It  is  seen  from  Fig.  329  that  the  hanger  and  the  diagonal  (at  C71) 
have  the  longest  lever  arms  and  hence  most  likely  the  pin  will  be  sub- 
jected to  the  maximum  moment  when  these  members  have  about  the 
maximum  simultaneous  stress.  As  this  can  be  ascertained  only  by  trial, 
let  us  place  the  loads  as  shown  at  (a),  Fig.  333,  in  which  case  wheel  4 
is  at  B.  Taking  moments  about  J  (using  Table  A)  we  obtain  the  reaction 
263,100  Ibs.  at  A,  and  taking  moments  about  B  we  obtain  the  concentra- 
tion 19,200  Ibs.  at  A.  Further,  taking  moments  about  C  and  A  we 
obtain  the  concentration  75, GOO  Ibs.  at  B — all  due  to  Cooper's  £40  load- 
ing. Then  beginning  at  joint  A  and  drawing  the  stress  diagram  shown 
at  (b)  we  obtain  the  stresses  in  all  the  members  at  joint  b  (Ul)  due  to 
Cooper's  £40  loading  and  multiplying  these  by  f  $  we  obtain  the  live- 
load  stresses  shown  at  (a),  which  are  due  to  the  £50  loading.  Next, 
multiplying  each  of  these  stresses  given  at  (a)  (except  the  stress  in  the 
hanger)  by  300+ (225  + 300)  we  obtain  the  impact  stress  in  each 
member.  Then  adding  together  these  live-load  stresses  and  the  impact 
and  the  dead-load  stresses  (given  in  Fig.  328),  we  obtain  the  stresses 
shown  at  (c). 

There  is  some  question  as  to  how  much  impact  should  be  added 
to  the  hanger.  Should  L  in  the  impact  formula  be  taken  as  50  or  225? 
As  is  seen,  the  impact  really  added  is  just  sufficient  to  balance  up  the 
vertical  components  on  the  pin.  This  is  a  fair  average  for  the  impact 
and  seems  to  be  a  rational  value. 

The  stresses  given  at  (c)  are  resolved  graphically  into  horizontal 
and  vertical  components  as  shown.  The  horizontal  components  for  half 
of  the  pin  are  shown  at  (d)  and  the  vertical  components  for  half  of 
the  pin  are  shown  at  (e). 

Proceeding  as  outlined  in  Art.  83,  for  the  bending  moment  on  the 
pin  due  to  the  horizontal  components,  we  have 

Me  =  0 

Mt  =  0  +  245,000  x  {J  =  +168,400  inch  Ibs. 

Mg  =  +168,400+  (-98,500x2^)  =-34,700  inch  Ibs. 

Similarly,  for  the  bending  moment,  due  to  the  vertical  components, 
we  have 

Me  =  0 

Mt  =  0  +  305,000  x  -1 J  =  +209,700  inch  Ibs. 

Mg  =  209,700  +  (212,000  x  2^)  =  +646,900  inch  Ibs. 

Mb  =  646,900  +  (90,500  x  If)  =+793,900  inch  Ibs. 

As  the  members  are  symmetrically  arranged  in  reference  to  the  cen- 
ter line  of  the  chord,  the  moment  at  g,  due  to  the  horizontal  forces,  is 
constant  between  the  two  diagonals  and  hence  for  the  resultant  maxi- 
mum moment  at  h,  we  have 


M  =x/34,7002  +  793,900  2=  797,200  inch  Ibs. 

This,  as  is  readily  seen,  is  the  maximum  resultant  bending  moment 
on  the  pin  when  the  live  load  is  in  the  position  shown  at  (a),  Fig.  333. 


DESIGN  OF  SIMPLE  BAILEOAD  BEIDGES  469 

In  fact,  it  is  about  the  maximum  moment  that  can  occur  on  the  pin. 
Applying  (1),  Art.  83,  taking  /  as  25,000  Ibs.,  we  find  that  the  above 
moment  (79  7,200"  #)  requires  a  6J"  pin  and  hence  the  7"  pin  assumed 
is  about  the  correct  size  as  far  as  the  bending  moment  is  concerned. 

In  case  of  doubt,  in  any  case,  as  to  the  position  of  the  live  load,  the 
loading  can  be  placed  in  different  positions  and  the  moment  on  the  pin  cal- 
culated for  each  position  in  the  same  manner  as  shown  above  for  that  one 
position.  This  method  of  procedure,  as  is  evident,  is  tedious.  As  a  rule, 
the  position  of  the  live  load  can  be  ascertained  near  enough  (as  the  impact 
is  questionable)  by  mere  inspection. 

The  maximum  shear  on  the  pin  at  each  side  of  the  chord  is  about 
330,000  Ibs.,  which  is  the  resultant  of  the  stresses  in  the  hanger  and 
diagonal.  Then  for  the  maximum  shearing  stress  on  the  7"  pin  we  have 
330,000 -r  38.4  =  8,590  Ibs.  per  sq.  in. 

From  this  it  is  seen  that  the  7"  pin  is  amply  large  as  far  as  shear 
is  concerned,  12,000  Ibs.  being  permissible,  and  as  it  is  the  correct  size 
for  bending  it  will  be  used. 

The  details  of  the  portal  and  laterals  at  [71  are  considered  to  be 
self-explanatory.  The  number  of  rivets  required  in  each  connection  is 
obtained  by  developing  the  section  of  the  member  connected,  as  explained 
in  Art.  184. 

Joint  LI.  All  of  the  details  at  this  joint  are  practically  self-ex- 
planatory. The  pin  supports  only  the  weight  of  the  bottom  chord.  The 
pin  plates  on  the  bottom  chord  are  intended  solely  to  replace  the  metal 
cut  from  the  chord  by  the  pin  hole.  The  rivets  connecting  the  bottom 
of  the  hanger  to  the  lateral  plate  should  be  sufficient  to  transmit  the 
component  of  the  lateral  in  panel  LO-L~L  along  the  bottom  chord.  This 
component  is  about  105,000  Ibs.  This  requires 

105,000  --  6,000  =  17— J "  field  rivets. 
16  are  used. 

The  component  of  the  stress  in  the  same  lateral  along  the  floor 
beam  is  about  70,000  Ibs.  This  requires  about  12— J"  field  rivets.  12 
are  used. 

The  pin  plates  on  the  hanger  are  for  the  purpose  of  reinforcing  the 
hanger  for  bending.  The  bending  on  the  hanger  about  the  pin  (at  LI), 
which  is  due  to  the  longitudinal  component  of  the  stress  in  the  bottom 
lateral,  is  105,000  x  14.25  =  1,496,250"#.  The  moment  of  inertia  of  the 
two  channels  is  625.2,  and  the  moment  of  inertia  of  the  2  —  -f-%"  plates 
is  246.1,  making  a  total  moment  of  inertia  of  871.3.  Then  for  the 
maximum  fiber  stress  in  the  hanger  at  LI,  due  to  bending,  we  have 

/  =  (1,496,250  x  7)  -  871.3  =  12,020  Ibs.  per  sq.  in. 

From  this  it  is  seen  that  the  hanger  is  amply  strong  to  resist  the  bending 
at  LI. 

The  rivets  connecting  the  lug  angles  to  the  bottom  of  the  hanger 
should  be  sufficient  to  transmit  the  longitudinal  component  of  the  stress 
in  the  lateral.  These  rivets  are  in  double  shear  and  bearing  on  about 
•Jf"  of  metal.  For  the  number  of  J"  shop  rivets  required  in  double 
shear,  we  have 

105,000^  (7,200x2)  =  7.3,  say  8. 


470 


STRUCTURAL  ENGINEERING 


-486OOO* 


Co) 

»  -6370OO' 


J 


.(c) 


For  the  number  required  in  bearing  on  the  |f  "  metal,  we  have 

105,000  -=-17,060  =  6. 

As  is  seen,  8  are  used  —  the  number  required  for  shear. 

Joint  L2.  The  details  of  this  joint  are  shown  in  Fig.  330.  The 
calculations  for  the  end  of  the  bottom  chord  are  just  the  same  as  given 
above  for  the  other  end  of  that  member  (at  LO). 

The  pin  bearing  on  the  post  should  be  sufficient  to  transmit  the  ver- 
tical component  of  the  maximum  stress  in  the  diagonal  U1-L2.  This 
component  is  equal  to  about  300,000  #.  Then  for  the  required  bearing 
on  the  post,  assuming  a  7"  pin,  we  have 

300,000  -=-  (7  x  24,000)  =  1.78  ins. 

or  0.89"  on  a  side.  The  thickness  of  the  web  of  the  40*  channel  is 
0.52"  (about  ||").  This  with  the  Ty  pin  plate  makes  H"  or  0.96", 
which  is  about  ^\  more  than  needed,  but  as  counter-sinking  (as  a  rule) 

is  not  permitted  in  metal  less 
than  Ty  thick  the  •&"  plates 
shown  will  be  used. 

About  the  maximum  bend- 
ing moment  on  the  pin  will  occur 
when  the  bottom  chord  L2-L3 
has  maximum  stress.  (This  can 
be  determined  in  the  manner 
shown  above  for  joint  £71.)  The 
wheels  are  in  the  position  shown 
in  Fig.  322  when  this  stress 
occurs.  Taking  moments  about 
J  (Fig.  322)  with  wheel  7  at  C 
we  can  obtain  the  reaction  R  at 
A  and  taking  moments  about  B 
we  can  obtain  the  concentration 
at  A.  Then  the  stress  in  the 
bottom  chord  AB  can  be  obtained  quickly  by  analyzing  graphically  the 
joint  A.  Then  the  stress  in  chord  BC  is  known,  as  it  is  equal  to  the  stress 
in  chord  AB.  Next  the  stress  in  the  diagonal  bC  (U1-L2)  is  readily  de- 
termined, as  the  horizontal  component  of  its  stress  is  equal  to  the  difference 
between  the  stress  in  chord  BC  and  CD  and  as  the  vertical  component  of 
the  stress  in  the  diagonal  is  equal  to  the  stress  in  the  post,  we  can  readily 
determine  the  stress  in  the  post  and  thus  we  would  have  all  the  live-load 
stresses  in  the  members  at  L2  determined. 

After  determining  these  live-load  stresses,  the  impact  is  determined 
in  each  member  by  multiplying  the  live-load  stress  in  it  by 
300  -=-(225  +  300).  Then  by  adding  together  the  live-load  stress  and 
impact  and  the  dead-load  stress,  we  obtain  the  total  stress  in  each 
member  as  given  at  (a),  Fig.  334.  Two-thirds  of  a  panel  load  of  dead 
load  is  added  to  the  post  to  balance  up  the  vertical  component  of  the 
diagonal.  However,  this  should  really  be  added  to  the  post  as  the  pin 
is  below  the  floor  beam  connection  and  hence  most  of  the  weight  at  the 
joint  is  transmitted  to  the  lower  end  of  the  post. 


Fig.  334 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  471 

The  horizontal  components  of  the  stresses  on  one-half  of  the  pin 
are  shown  at  (b)  (Fig.  334),  and  the  vertical  components  on  one-half 
of  the  pin  are  shown  at  (c). 

For  the  bending  moment  on  the  pin,  due  to  the  horizontal  com- 
ponents, we  have 

Me  =  0 

Mt  =  0  +  (175,000  x  If)  =  306,250  inch  Ibs. 

Mg  =  +306,250  +  (-68,000  x  2|  )  =  +144,750  inch  Ibs. 

Mh  =  +144,750  +  (  100,500  x  1  j)  =  +295,500  inch  Ibs. 

For  the  bending  moment  on  the  pin,  due  to  the  vertical  component, 
we  have 


Mk  =  0  +  (124,000  x  1TV  )  =  178,200  inch  Ibs. 
Then  for  the  maximum  resultant  moment  on  the  pin,  we  have 


M  =  J295,5002  + 178,2002  =  345,000  inch  Ibs. 

Then  by  applying  (1),  Art.  83,  taking  /  as  25,000,  we  find  that  the 
above  moment  (  345,000"  #)  calls  for  a  5J"  pin  (about). 

The  maximum  shear  on  the  pin  is  about  175,000  Ibs.,  which  requires 
that  the  pin  be  only  about  4  j^''  diameter. 

From  the  above  it  is  seen  that  the  assumed  7"  pin  is  larger  than 
necessary,  but  the  7"  will  be  used,  for  little  would  be  saved  by  reducing 
the  size,  as  the  thickness  of  pin  bearing  on  the  bottom  chord  (L2-LO) 
and  also  on  the  post  U2-L2  would  have  to  be  increased  if  a  smaller  pin 
were  used. 

The  other  details  at  L2  are  readily  understood.  The  calculation 
of  the  details  of  the  bottom  laterals,  shown  at  this  joint,  is  mostly  a 
matter  of  developing  the  members.  The  rivets  connecting  the  bottom 
of  the  post  to  the  lateral  plate  should  be  sufficient  to  transmit  the  longi- 
tudinal component  of  the  stress  in  the  lateral  in  panel  L1-L2. 

Joint  U2.  The  top  chord  has  a  butt  joint  at  this  point;  that  is, 
the  ends  of  the  members  are  planed  so  that  they  bear  tightly  against 
each  other  and  hence  the  stress  is  transmitted  from  one  chord  member 
to  the  other  chord  member  without  passing  through  the  pin.  The  splice 
plates  on  the  chord  are  intended  only  to  hold  the  chords  in  line.  How- 
ever, there  must  be  sufficient  pin  bearing  on  the  chord  to  take  the  com- 
ponent (along  the  chord  U2-U3)  of  the  stress  in  diagonal  U2-L3.  The 
maximum  stress  in  the  diagonal  (see  Fig.  328)  is  280,000  Ibs.  Resolv- 
ing this  along  the  chord  U2-U3  (which  can  be  done  graphically)  we 
obtain  a  component  of  159,000  Ibs.  To  transmit  this  component,  assum- 
ing a  6"  pin,  requires  159,000+  (24,000  x  6)  =1.1//  thickness  of  bearing, 
or  about  \n  on  each  side  of  the  chord.  As  is  seen,  there  are  2  —  f" 
plates  and  a  TV'  web>  making  in  all  1-&"  bearing  on  each  side  of  the 
chord,  which  is  quite  excessive. 

The  pin  bearing  on  the  post  should  be  sufficient  to  transmit  the 
maximum  stress  in  the  post,  which  is  221,000  Ibs.  (See  Fig.  328.) 
This  requires  a  bearing  (assuming  a  6"  pin)  of  221,000-=-  (24,000x6) 


472  STRUCTURAL  ENGINEERING 

=  1.54"  or  0.72"  on  a  side.  The  web  of  the  40#  channel  is  ft"  thick. 
This  with  the  •{%"  plate  makes  j-j-"  thickness  of  bearing  on  each  side  of 
the  post,  which  is  excessive. 

The  maximum  bending  moment  on  the  pin  will  occur  when  the 
diagonal  (J72-Z/3)  has  maximum  stress,  as  the  post  (U2-L2)  has  about 
the  maximum  stress  at  the  same  time.  By  placing  the  live  load  for 
maximum  stress  in  the  diagonal  (as  shown  in  Fig.  315)  and  determin- 
ing the  reaction  at  A  and  the  concentration  at  C  (these  are  given  on 
pages  432  and  433)  the  live-load  stress  in  each  of  the  members  at  U2 
can  be  readily  determined  by  graphics.  Then  determining  the  impact 
by  multiplying  each  stress  by  300 -r  (150  +  300)  and  adding  together 
this  impact  and  the  live-  and  dead-load  stresses  and  resolving  the 
resulting  stresses  horizontally  and  vertically,  the  maximum  bending 
moment  on  the  pin  can  be  determined  in  the  same  manner  as  shown  above 
for  the  other  joints.  This  maximum  moment  is  about  190,000  inch  Ibs. 
This  requires  about  a  4J"  pin.  From  the  above,  it  is  seen  that  the  6" 
pin  is  larger  than  theoretically  required.  But  it  is  usual  practice  to 
limit  the  minimum  diameter  of  a  pin  to  -f^  of  the  width  of  the  smallest 
eye-bar  connected.  The  eye-bars  at  U2  (as  seen)  are  7"  wide.  So, 
according  to  this  requirement,  the  pin  should  be  5.6"  in  diameter  and 
hence  the  assumed  6"  pin  is  about  the  correct  size,  and  will  be  used. 
The  other  details  at  U2  are  readily  understood. 

The  details  at  £73  are  similar  to  those  at  U2  and  the  details  at  L3 
are  very  similar  to  those  at  L2  and  hence  the  calculations  of  the  details 
at  these  joints  are  quite  similar  to  those  given  above  and  consequently 
are  omitted.  The  same  can  be  said  regarding  joints  £74  and  1/4  (Fig. 
331).  The  details  at  these  joints  are  readily  understood. 

The  calculations  of  the  details  of  the  intermediate  floor  beams, 
shown  in  Fig.  331,  are  almost  exactly  as  given  in  Art.  186  (pages  410 
to  412)  for  the  floor  beams  in  the  150- ft.  riveted  span. 

After  the  general  drawings  (Figs.  329,  330  and  331)  are  completed, 
the  shop  drawings  and  shop  bills  for  the  structure  can  be  made  by 
proceeding  in  the  same  manner  as  outlined  in  Art.  186  for  the  150- 
ft.  span. 

204.  Camber. — The  camber  of  the  trusses  in  this  case  is  obtained 
by  lengthening  the  top  chord  y  for  each  10  feet  of  horizontal  length, 
as  explained  in  Art.  185.  The  corresponding  lengths  of  the  diagonals 
are  calculated  by  using  the  mean  lengths  of  the  top  and  bottom  chords 
in  each  case.  For  example,  the  horizontal  uncambered  length  of  top 
chord  U2-U3  (Fig.  330)  is  25  ft.  So  the  cambered  horizontal  length 
would  be  25'  -  0-jffr".  Taking  this  as  the  base  of  a  right  angle  triangle, 
and  the  difference  between  the  lengths  of  posts  U3-L3  and  U2-L2  as  the 
altitude,  the  cambered  length  of  the  top  chord  U2-U3  (given  as  25'  -5J") 
is  computed  as  the  hypotenuse  of  the  triangle.  Then  taking  25'  -  O^" 
as  the  base  of  a  right  angle  triangle  and  the  length  of  post  U2-L2  as  the 
altitude,  the  length  of  diagonal  U2-L3  (given  as  45'  -  4£")  is  calculated 
as  the  hypotenuse  of  that  triangle.  The  length  of  the  end  post  is  not 
increased. 

The  determination  of  actual  amount  of  camber  of  bridge  trusses  and 
also  the  determination  of  the  theoretical  camber  of  the  same  will  be 
given  later. 


DESIGN  OF  SIMPLE  KAILKOAD  BRIDGES 


473 


205.  Graphical  Determination  of  Live-Load  Stresses  in  Curved 
Chord  Pratt  Trusses. — The  influence  line  method,  outlined  in  Art.  103,, 
is  the  most  convenient  graphical  method  to  use  in  this  case,  as  the  stresses 
in  the  truss  members  are  thus  determined  directly  without  considering 
either  shears  or  moments. 

Chord  Members.  Let  it  be  required  to  determine  the  live-load  stress 
in  the  top  chord  cd  of  the  truss  shown  in  Fig.  335,  due  to  Cooper's 
-E40  loading. 

It  is  obvious  that  a  single  load  moving  from  J  to  the  left  over  the  span 
will  cause  a  stress  in  chord  cd  which  will  be  zero  when  the  load  is  at  J  and 


Fig.  335 

increase  constantly  until  the  load  reaches  point  D  and  that  this  stress  will 
then  decrease  constantly  as  the  load  moves  on  to  the  left  from  D,  becoming 
zero  when  the  load  reaches  A.  Then  evidently  the  influence  line  for 
the  stress  in  chord  cd  is  of  the  form  c'o'd' ,  shown  at  (c).  Now,  know- 
ing the  form  of  the  influence  line,  the  next  thing  is  to  construct  it.  This, 
as  is  readily  seen,  is  a  simple  matter  after  the  ordinate  o'g'  is  known. 
This  ordinate  o' 'g' ' ,  as  we  know,  is  the  stress  in  the  top  chord  cd,  due 
to  a  unit  load  at  D.  Then,  of  course,  the  first  thing  to  do  in  construct- 
ing the  desired  influence  line  is  to  determine  the  stress  in  the  top  chord 
cd,  due  to  a  unit  load  at  D.  This  stress  can  be  determined  very  readily 


474  STRUCTURAL  ENGINEERING 

either  analytically  or  graphically.  The  graphical  determination  is  the 
more  convenient  and  hence  that  method  will  be  used.  By  drawing,  at 
(a),  the  lines  00',  OD  and  O'D  we  obtain,  as  we  may  say,  the  truss 
ODO'  ,  which  will  have  end  reactions  exactly  equal  to  those  of  the  bridge 
truss,  due  to  a  unit  load  at  D,  and  as  the  member  00'  coincides  with 
top  chord  cd,  the  stress,  due  to  this  unit  load  at  D,  will  be  the  same 
in  the  two.  This  is  readily  seen,  for,  taking  moments  about  D,  we 
obtain 


for  the  stress  in  each,  where  R  represents  the  reaction  at  either  A  or 
0  due  to  a  unit  load  at  D. 

By  drawing  mp  (at  (b)  )  and  laying  off  nm  (equal  to  one  inch) 
equal  to  one  pound  and  drawing  np,  we  have  the  influence  line  for  the 
reactions  at  A.  Then  gh  is  equal  to  the  reaction  at  A  and  also  at  0,  due 
to  a  unit  load  at  D. 

Then  by  drawing  from  h  a  line  parallel  to  OD  and  from  g  a  line 
parallel  to  00'  ,  we  obtain  the  line  og,  which  is  equal  (in  inches)  to  the 
stress  in  chord  cd,  due  to  the  unit  load  at  D.  In  other  words,  by  con- 
structing a  diagram  of  the  forces  at  0,  we  obtain  the  stress  in  00', 
which  is  equal  to  the  stress  in  the  chord  cd,  due  to  a  unit  load  at  D. 

Then  by  drawing  c'd'  (at  (c)  )  and  laying  off  o'g'  —og  and  draw- 
ing c'o'  and  o'd't  we  have  the  desired  influence  line  for  the  stress  in  top 
chord  cd. 

Then  placing  the  loading  for  the  maximum  moment  about  D,  accord- 
ing to  Art.  91  (see  Fig.  323),  and  drawing  the  ordinates  1,  2,  ....  7,  as 
shown,  and  multiplying  each  by  the  load  or  group  of  loads  at  each,  and 
adding  together  the  results,  the  desired  maximum  stress  in  top  chord  cd 
will  be  obtained. 

Ordinates  1  and  2  are  drawn  at  the  center  of  gravity  of  equal 
groups.  Then  by  adding  together  the  ordinates  1  and  2  and  multiply- 
ing by  the  weight  of  one  of  the  groups,  the  stress  in  top  chord  cd,  due 
to  the  two  groups,  will  be  obtained.  Similarly,  by  adding  together  the 
ordinates  3  and  4  and  multiplying  by  the  weight  of  one  of  the  groups 
(at  either  of  the  ordinates),  the  stress  in  the  top  chord  cd,  due  to  those 
two  groups,  will  be  obtained.  Next,  adding  together  ordinates  5  and  6 
and  multiplying  by  the  weight  of  one  of  the  wheels,  the  stress  in  top 
chord  cd,  due  to  the  two  wheels  (1  and  10),  will  be  obtained.  Multiply- 
ing the  ordinate  7  (which  is  equal  to  one-half  of  ordinate  8)  by  the  total 
uniform  load  the  stress  in  chord  cd,  due  to  the  uniform  load,  is  obtained. 
Then  adding  together  all  of  these  results,  the  total  maximum  live-load 
stress  in  top  chord  cd  is  obtained. 

The  combined  length  of  ordinates  1  and  2  (which  can  be  quickly 
combined  by  the  use  of  a  pair  of  dividers,  as  shown  in  Example  1  of 
Art.  100)  scales  1.54  ins.  This  means  that  if  (as  wm  =  l"  =  l#)  one 
unit  load  (1  Ib.)  be  placed  on  the  truss  exactly  over  ordinate  1  and 
another  exactly  over  ordinate  2,  these  two  unit  loads  would  produce  a 
stress  of  1.54  Ibs.  in  top  chord  cd.  Then  evidently  the  two  groups  of 
wheels  at  these  ordinates,  each  weighing  80,000  Ibs.,  will  produce  a  stress 


DESIGN  OF  SIMPLE  EAILKOAD  BEIDGES  .        475 

in  the  top  chord  cd  of 

1.54  x  80,000  =  123,000  Ibs. 

The  combined  length  of  ordinates  3  and  4  scales  1.73  ins.  Then 
for  the  stress  in  top  chord  cd  due  to  the  two  groups  of  wheels  at  these 
ordinates,  we  have 

1.74  x  52,000  =  90,480,  say  90,500  Ibs. 

Similarly  for  the  stress  due  to  wheels  1  and  10,  the  combined  length 
of  ordinates  5  and  6  being  1.23  ins.,  we  have 

1.23x10,000  =  12,300  Ibs. 

Ordinate  7  scales  0.41  ins.  Then  for  the  stress  in  top  chord  cd, 
due  to  the  uniform  load,  we  have 

0.41  x  (2,000  x  105)  =  86,100  Ibs. 
Now,  adding  together  the  above,  we  obtain 
123,000  +  90,500  +  12,300  +  86,100  =  311,900,  say,  312,000  Ibs. 

for  the  maximum  stress  in  top  chord  cd  due  to  Cooper's  £40  loading, 
and  multiplying  this  by  f$  we  obtain  390,000  Ibs.  for  the  stress  due  to 
the  £50  loading.  As  is  seen  from  Fig.  328,  this  is  6,000  Ibs.  less  than 
the  actual  stress,  but  as  this  difference  is  less  than  2  per  cent,  the  result 
is  accurate  enough.  That  is,  the  change  of  section  of  the  top  chord 
resulting  from  the  error  would  be  negligible. 

As  the  horizontal  component  of  the  stress  in  top  chord  cd  is  equal 
to  the  stress  in  bottom  chord  DE,  as  previously  shown,  and  the  position 
of  the  loading  for  maximum  stress  in  the  two  members  being  the  same, 
the  influence  line  for  the  stress  in  bottom  chord  DE  can  be  quickly 
constructed  after  the  one  for  the  top  chord  cd  is  completed.  Thus:  By 
drawing  or  we  obtain  the  stress  in  bottom  chord  DE,  due  to  a  unit 
load  at  D.  Then  drawing  D'E'  (at  (d)  )  and  laying  off  r'o" '  —  or  and 
drawing  D'r'  and  r'E',  we  obtain  the  influence  line  for  the  stress  in 
chord  DE.  The  stresses  can  then  be  determined  in  this  member  in  the 
same  manner  as  shown  above  for  top  chord  cd.  However,  the  stress 
in  bottom  chord  DE  can  be  obtained  more  readily  as  the  horizontal 
component  of  the  stress  in  top  chord  cd.  The  method,  of  course,  can 
be  reversed;  that  is,  the  influence  line  for  the  stress  in  bottom  chord 
DE  could  be  constructed  and  the  stress  in  DE  determined  and  then  the 
stress  in  top  chord  cd  could  be  obtained  by  multiplying  the  stress  found 
in  DE  by  the  secant  of  the  slope  angle  of  chord  cd. 

The  construction  of  the  influence  lines  and  determination  of  the 
stresses  in  the  other  chord  members  are  practically  the  same  as  shown 
above  for  chords  cd  and  DE. 

For  example,  by  drawing  0"0f",  0"E  and  E0"f  and  constructing 
a  diagram  at  ordinate  z  of  the  forces  at  0" ,  the  stress  in  top  chord  de, 
due  to  a  unit  load  at  E,  is  obtained.  Then  the  influence  line  for  the 
stress  in  top  chord  de,  as  well  as  for  bottom  chord  EF,  is  readily 
constructed. 

As  another  example,  by  drawing  tv   (at   (b)  )   parallel  to  the  end 


476  STRUCTURAL  ENGINEERING 

post,  we  obtain  vu,  which  is  equal  to  the  stress  in  bottom  chord  AB  (also 
BC},  due  to  a  unit  load  at  B.  Then  drawing  A'B'  (at  (e)  )  and  laying 
off  v'u* '  =  vu  (which  can  be  done  by  using  a  pair  of  dividers)  and  drawing 
A'v'  and  v'B'}  we  obtain  the  influence  line  for  the  stress  in  bottom  chords 
AB  and  EC,  shown  at  (e). 

Web  Members.  As  an  example,  let  it  be  required  to  determine  the 
stress  in  diagonal  cD.  The  influence  line  for  the  stress  in  this  member,, 
as  explained  in  Art.  103,,  will  be  of  the  form  A's'o'B' ',  shown  at  (c)  in 
Fig.  336.  To  construct  this  influence  line,  first  draw  the  influence  lines 
for  reactions,  as  shown  at  (b).  Then  gh  represents  the  reaction  at  A, 
due  to  a  unit  load  at  D.  Next,  draw  00',  OD  and  O'D.  Then,  drawing 
og  and  oh  (at  (b)  )  parallel,  respectively,  to  00'  and  OD,  and  ko 
parallel  to  diagonal  cD,  we  have  the  stress  in  the  diagonal  cD,  due  to  a 
unit  load  at  D,  represented  by  this  line  ko.  Then,  by  drawing  A'B' 
(at  (c)  )  and  laying  off  h'o'  equal  to  ko,  the  part  o'B'  of  the  influence 
line  can  be  drawn. 

Next,  draw  01  and  /DO',  as  shown  at  (a).  Then,  drawing  vs  and 
su  parallel,  respectively,  to  10'  and  00'  and  is  parallel  to  the  diagonal 
cD,  we  have  the  stress  in  the  diagonal  cD,  due  to  a  unit  load  at  C,  repre- 
sented by  this  line  ts.  Then,  by  laying  off  t's'  (at  (c)  )  equal  to  is, 
the  parts  s' 'A'  and  s'o'  of  the  influence  line  can  be  drawn,  which  will 
complete  the  construction  of  the  influence  line  A's'o'B'. 

Any  load  at  any  point  to  the  right  of  N  will  produce  tension  in 
diagonal  cD,  and  any  load  to  the  left  will  produce  compression  in  the 
same.  Then,  evidently,  to  obtain  the  maximum  tension  in  the  diagonal, 
all  the  loads  should  be  placed  to  the  right  of  N.  According  to  Art. 
190,  a  wheel  will  be  at  D  when  the  maximum  tension  in  cD  occurs.  Then, 
evidently,  the  loading  will  be  in  the  position  for  maximum  stress  in 
diagonal  cD  if  the  wheel  be  placed  at  &'  that  brings  wheel  1  the  closest 
to  N,  the  limit  being  that  wheel  1  should  not  be  to  the  left  of  N.  From 
this  it  is  seen  that  the  position  of  the  loading  for  maximum  stress  in 
the  diagonal  can  be  determined  directly  from  the  influence  line  instead 
of  applying  the  criterion  of  Art.  190. 

The  position  of  the  loading  for  maximum  tension  in  diagonal  cD  is 
shown  at  (c).  By  multiplying  the  ordinates  by  the  loads,  as  explained 
above,  the  maximum  tension  in  diagonal  cD  will  be  obtained. 

In  any  case  when  a  group  of  wheels  comes  at  a  point  where  the 
influence  line  changes  direction,  as  wheels  2  ...  5  do  in  this  case,  the  ordi- 
nate  at  the  center  of  gravity  of  the  group  can  not  be  used.  In  such 
cases  the  weight  of  each  wheel  can  be  multiplied  by  the  ordinate  at  it, 
or  the  ordinates  at  the  center  of  gravity  of  the  wheels  to  either  side  of 
the  change  of  slope  of  the  influence  line  can  be  used,  or  the  sum  of  all 
of  the  ordinates  can  be  obtained  by  the  use  of  dividers  (see  Example 

1,  Art.  100),  and  this  sum  multiplied  by  the  weight  of  one  wheel.     Thus, 
in  the  case  shown  at  (c)  the  tensile  stress  in  diagonal  cD,  due  to  wheels 
2  ...  5,  can  be  obtained  by  multiplying  ordinate  1  by  the  weight  of  wheel 

2,  and  ordinate  3  by  the  weight  of  wheels  3,  4  and  5,  or  by  multiplying 
the  sum  of  all  the  ordinates  1,  2,  3  and  4  by  the  weight  of  one  wheel. 

The  maximum  compression  in  diagonal  cD  can  be  obtained  by  revers- 
ing the  loading;  that  is,  bringing  it  on  from  the  left,  and  placing  the 
wheel  at  t'  that  brings  wheel  1  closest  to  N  without  passing  that  point. 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES 


477 


The  actual  stress  Is  then  obtained  by  multiplying  the  loads  by  the  ordi- 
nates,  as  previously  explained. 

The  influence  line  for  the  maximum  compression  in  post  dD  is 
shown  at  (d).  To  obtain  this  influence  line  00',  OD,  O'D,  ODI'  and 
O'lf  should  be  drawn.  Then,  by  drawing  fy  parallel  to  00' ,  yl  parallel 
to  01',  and  yz  parallel  to  post  dD,  we  have  the  stress  in  post  dD,  due 
to  a  unit  load  at  E,  represented  by  yz.  Then,  drawing  C'D'  at  (d), 
and  laying  off  y' z'  equal  to  yz,  we  can  draw  the  part  y'D'  of  the  influ- 
ence line.  Next,  drawing  qw  parallel  to  00' ,  wh  parallel  to  O'D,  and 
ew  parallel  to  post  dD,  we  have  the  stress  in  post  dD,  due  to  a  unit  load 
at  D,  represented  by  line  ew.  Then,  laying  off  e'wf ',  at  (d),  equal  to  ew, 
we  can  complete  the  influence  line  C'w'y'D'  by  drawing  C'w'  and  w' y' . 


Fig.  336 


The  maximum  compression  in  post  dD  can  then  be  determined  by 
placing  the  loading  on  PD'  with  the  wheel  at  z'  that  brings  wheel  1 
the  closest  to  P,  and  proceeding  as  explained  above  in  the  case  of  the 
diagonal  dD.  Influence  lines  for  maximum  tension  in  the  posts  will 
be  considered  later. 

The  influence  lines  for  maximum  stress  in  the  other  diagonals  and 
posts  and  determination  of  the  stresses  are  similar  to  the  above. 

The  influence  lines  can  be  constructed  more  readily  in  the  following 
manner,  taking  first  the  one  shown  at  (c)  in  Fig.  336: 

The  ordinate  o'k'  can  be  obtained  in  the  manner  explained  above, 


478 


STRUCTURAL  ENGINEERING 


and  point  N  can  be  located  by  drawing  OCx  and  O'Dx,  as  x  is  directly 
over  N.  Having  the  ordinate  o'k'  determined  and  point  N  located,  the 
influence  line  A's'o'B'  can  be  drawn.  The  ordinate  #V,  shown  at  (d), 
can  be  obtained  in  the  manner  explained  above,  and  point  P  can  be 
located  by  drawing  ODx'  and  O'Ex* ',  as  x'  is  directly  over  P.  Having 
the  ordinate  y'z'  determined  and  the  point  P  located,  the  influence  line 
C'w'y'D'  can  be  drawn. 

A  better  way  to  locate  points  N  and  P  is  to  draw  the  line  OJ. 
Then  point  F,  where  OJ  intersects  the  diagonal  cD,  is  directly  over 
point  N;  and  point  G,  where  the  line  OJ  intersects  diagonal  dE,  is 
directly  over  point  P. 

It  can  be  readily  shown  that  this  construction  is  correct.  Referring 
to  Fig.  337,  the  structure  OCDO'cO,  supporting  the  loads  PI  and  P2, 
is  a  rigid  structure.  It  is  evident  that  the  loads  PI  and  P2  could  have 


Fig.  337 

such  relative  weights  that  the  member  cD  (also  cC)  would  have  zero 
stress.  In  that  case,  OCDO'  would  be  an  equilibrium  polygon.  Then 
by  prolonging  the  segments  O'D  and  OC  to  intersection,  the  point  a? 
is  obtained,  which  would  be  at  the  resultant,  or  center  of  gravity,  of 
the  loads  PI  and  P2  (see  Art.  45).  Then  both  of  the  loads'  could 
be  placed  at  x  and  the  stress  in  member  cD  (also  cC)  would  be 
zero.  Now,  it  is  evident  that  as  long  as  PI  and  P2  have  this  same 
relative  value,  OCDO'  would  remain  an  equilibrium  polygon,  regardless 
of  the  actual  weights  of  the  loads  PI  and  P2;  so,  evidently,  the  stress 
in  cD  due  to  any  load  at  x  will  be  zero,  and  hence  x  is  the  point  of 
zero  stress  for  diagonal  cD. 

From  similar  triangles  Dmx  and  DJO',  we  have* 


from  which  we  obtain 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  479 

From  similar  triangles  OAC  and  xmC,  we  have 

£  =  -, 
iTom  which  we  obtain 


Now  equating  (1)   and  (2)  equal,  we  have 

6A  =  a    ' 
from  which  we  obtain 

0  _  ^  f*\ 

~~   —     77 10  )  . 

c        o/c 
From  similar  triangles  cFT  and  DFU,  we  have 

(4). 

But  from  similar  triangles  O'JO  and  cOT,  we  have 


y_a 
from  which  we  obtain 


h~L' 


And  from  similar  triangles  JAO  and  JDU,  we  have 


from  which  we  obtain 


Now  substituting  these  values  of  ?/  and  z  in  (4),  we  obtain 

--— 
t   ~bk' 

Equating  (3)  and  (5)  equal,  we  obtain 

&r.  ................  •  ............  • 

But  c  =  d-g  and  t  =  d-s. 

So,  substituting  these  values  in  (6)   and  reducing,  we  obtain 


which  proves  that  points  F  and  a  are  on  the  same  vertical  line. 

Any  of  the  other  cases  can  be  proved  in  a  similar  manner.     Refer- 
ring to  Fig.  336,  point  1  is  the  point  of  zero  stress  for  diagonal  bC  and 


480 


STRUCTURAL  ENGINEERING 


point  2  is  the  point  of  zero  stress  for  post  cC.  Point  3  is  the  point  of 
zero  stress  for  diagonal  dE  and  point  4  is  the  point  of  zero  stress  for 
post  eE.  The  point  of  zero  stress  for  diagonal  eF  is  at  5. 

Tension  in  Posts.  Let  it  be  required  to  determine  the  maximum  live- 
load  tensile  stress  in  post  dD  (Fig.  338).  First  construct  the  influence 
line,  shown  at  (c),  for  stress  in  diagonal  dE.  As  explained  on  pages 
424  and  425  (Art.  190),  the  maximum  live-load  tension  will  occur  in 
post  dD  when  the  load  is  so  placed  that  the  stress  in  diagonal  dE  is 
zero.  The  stress  in  the  diagonal  dE  due  to  dead  load  is  tension.  A 
live  load  moving  onto  the  bridge  from  the  left  would  begin  reversing 
this  dead-load  tension  from  the  start  and  continue  reversing  it  until  a 


Fig.  338 


point  was  reached  where  the  stress  in  the  diagonal  would  be  zero  and 
from  there  on,  as  the  load  moved  on  to  the  right,  the  stress  in  the  diagonal 
would  be  compression  and  this  compression  would  continue  to  increase 
until  the  live-load  extended  from  M  to  N  (see  the  influence  line  at  (c)  ) 
at  which  time  the  live-load  compression  would  be  a  maximum.  Then  as 
the  load  moved  on  past  N  the  compression  in  the  diagonal  would  decrease 
until  a  point  was  reached,  as  the  load  moved  on  to  the  right,  where  the 
stress  in  the  diagonal  dE  would  be  zero  for  the  second  time.  This  last 
position  is  the  one  desired,  as  the  maximum  tension  in  post  dD  will 
occur  when  the  load  is  in  that  position.  The  first  thing  is  to  find  that 
position. 

Let  S  represent  the  dead-load  tension  in  diagonal  dE  and  S'  the 
live-load  compression  in  the  same  when  the  load  extends  from  M  to  N. 
Then  the  load  to  the  right  of  N  must  be  just  sufficient  to  produce  a  ten- 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  481 

sion  of  S'  —  S  pounds  in  diagonal  dE.  An  equivalent  uniform  live  load 
of  w  pounds  per  foot  of  truss  will  be  used.  Let  y  be  the  distance  from 
H  to  the  head  of  the  uniform  live  load  and  let  A  represent  the  required 
shaded  area.  Then  the  required  tension  in  the  diagonal  is  Aw,  which 
must  be  equal  to  *S"  -  S,  so  we  have 

Aw  =  S'-S, 
from  which  we  obtain 

S'-S 


for  the  required  shaded  area  in  square  feet.     The  next  thing  is  to  find 
the  value  of  y  when  the  shaded  area  is  equal  to  (S'  —  S)/w. 
Referring  to  the  influence  line  at  (c),  we  have 


(7). 

Cut    ..'••  ,=K. 

So,  substituting  this  value  of  z,  and  ($'-£)  /w  for  A  in  (7),  and  reduc- 
ing, we  obtain 

-s)  ............................  (8). 


From  this  equation  (8)  the  head  of  the  equivalent  uniform  live 
load  can  be  located.  Then,  having  the  load  located,  the  influence  line 
for  the  maximum  live-load  tension  in  the  post  dD  can  be  constructed 
as  shown  at  (d),  and  by  multiplying  the  shaded  area  by  w  the  desired 
live-load  tension  in  the  post  will  be  obtained. 

To  construct  the  influence  line  shown  at  (d)  :  First  assume  zero 
stress  in  diagonal  dE.  Placing  a  unit  load  at  D  and  drawing  kh  parallel 
to  O'D  and  kg  parallel  to  O'd,  we  have  the  stress  in  top  chord  cd  given 
by  the  line  kg;  and  as  the  horizontal  components  of  top  chords  cd 
and  de  must  be  equal  (since  the  stress  in  dE  is  zero)  by  drawing  from 
g  a  line  parallel  to  top  chord  de  intersecting  the  vertical  mk  at  m,  we 
have  the  stress  in  chord  de  represented  by  the  line  gm  and  the  stress  in 
the  post  dD  represented  by  the  line  mk.  Then  laying  off  m'k'  at  (d) 
equal  to  mk  the  influence  line  as  shown  can  be  constructed. 

In  case  the  uniform  live  load  does  not  extend  to  ordinate  e,  as  shown 
at  (e),  it  is  better  to  locate  the  head  of  the  load  with  reference  to  P. 
Let  x  represent  the  distance  from  P  to  the  head  of  the  uniform  live  load. 
Then  we  have 

.    ce        2     „ 
A:  —  ::x2:c2, 


2 
from  which  we  obtain 


(9). 


206.    Determination  of  Dead-Load  Stresses  in  Pettit  Trusses.— 
Let  it  be  required  to  determine  the  dead-load  stresses  in  the  truss  shown 


482 


STRUCTURAL  ENGINEERING 


at  (a),  Fig.  339.  Let  JF1,  W2,  and  so  on  (all  of  which  are  equal) 
represent  the  loads  at  the  lower  panel  points;  PI,  P2,  and  so  on  (all 
of  which  are  equal)  represent  the  loads  at  the  upper  panel  points;  and 
let  R  represent  the  reaction  due  to  these  loads. 

Members  Ul-LO,  LO-L1,  L1-L2  and  U1-L1.  As  previously  shown 
(Art.  189),  the  dead-load  stress  in  Ul-LO  is  equal  to  Rsec6,  RtanB  in 
both  LO-Ll  and  L1-L2,  and  Wl  in  Ul-Ll. 

Member  U1-U2.  The  stress  in  this  member  is  readily  obtained  by 
taking  moments  about  L2.  Thus,  we  obtain 


for  the  horizontal  component  of  the  stress  in  U1-U2.     The  stress  in  the 
member  is  then  obtained  by  multiplying  H  by  sec</>. 


Fig.  339 


Member  U1-L2.  Having  determined  H,  the  horizontal  component 
of  the  stress  in  top  chord  U1-U2,  the  vertical  component  of  the  same 
is  equal  to 


Now,  considering  the  part  of  the  structure  to  the  left  of  section  1-1  and 
summing  up  the  vertical  forces,  we  obtain 


for  the  vertical  component  of  the  stress  in  U1-L2.     Then  we  have 
for  the  stress  in  U1-L2. 

The  stress  in  this  diagonal  can  also  be  determined  by  taking  moments 
about  the  intersection  of  the  chords  as  explained  in  Art.  189. 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  483 

Members  M3-L2  and  M3-L3.  In  determining  the  stresses  in  these 
two  members  the  triangle  L2-M3-L4:  can  be  considered  as  a  separate 
truss,  as  shown  at  (b).  The  stress  in  M3-L2  is  equal  to  ^(P3  +  W3)secO', 
and  the  stress  in  the  hanger  M3-L3  is  simply  equal  to  W3. 

Members  U2-L2.  Considering  the  part  of  the  structure  to  the  left 
of  section  2-2  shown  at  (c)  and  resolving  the  stress  in  L2-M3  both 
vertically  and  horizontally  at  L2  and  resolving  the  stress  in  U1-U2  and 
adding  up  the  vertical  forces,  we  have 


from  which  we  obtain 


for  the  dead-load  stress  in  post  U2-L2.  This  stress  can  also  be  deter- 
mined by  taking  moments  about  0.  Thus,  taking  moments  about  0,  we 
have 


from  which  we  obtain 


s  +  Zd 

for  the  stress  in  U2-L2. 

Member  M2-M3.  This  member  is  not  subjected  to  any  direct  stress. 
It  is  for  the  purpose  of  stiffening  post  U2-L2. 

Member  U2-U3-U4-  Considering  the  part  of  the  structure  to  the 
left  of  section  4-4,  as  shown  at  (d),  and  by  taking  moments  about  Z/4, 
we  obtain 


for  the  horizontal  component  of  the  stress  in  top  chord  U2-U3-U4:.    Then 
by  multiplying  J/l  by  sec</>'  the  stress  in  the  member  is  obtained. 

Member  M3-L4.  Having  HI,  the  horizontal  component  of  the  stress 
in  the  top  chord  U2-U3-U4:  determined,  the  vertical  component  F2  is 
equal  to  H\  tan<£'.  Then  resolving  the  stress  in  M3-L4  vertically  and 
horizontally  and  summing  up  the  vertical  forces  shown  at  (d)  we  obtain 

F3  =  R  -  (Wl  +  W2  +  W3)  -  (PI  +  P2  +  P3)  -  72 

for  the  vertical  component  of  the  stress  in  diagonal  M3-L4.     Then  for 
the  stress  in  the  member  we  have 


Member  L2-L3-LJ^.  Considering  the  part  of  the  structure  shown 
at  (d),  and  considering  S2  and  S3  instead  of  their  components  (see  Art. 
41),  and  taking  moments  about  U2  of  the  forces  shown  at  d,  we  obtain 


for  the  stress  in  bottom  chord  L2-L3-L4. 


484  STRUCTURAL  ENGINEERING 

Member  U2-M3.  The  vertical  component  of  M3-L2  is  equal  to 
J(P3  +  ?F3);  the  vertical  component  of  the  top  chord  U2-U3-U4,  as 
found  above,  is  equal  to  F2.  Then  considering  the  part  of  the  structure 
to  the  left  of  section  3-3  and  summing  up  the  vertical  forces  and 
components  on  same,  we  obtain 

F4  -  R  -  (  Wl  +  W2  +  PI  +  P2  )  -  J  (P3  +  W3  )  -  F2 

for  the  vertical  component  of  the  stress  in  diagonal  U2-M3.  Then  the 
stress  is  equal  to  F4  x  sec#'. 

Member  U3-M3.  This  member  simply  supports  the  load  P3  and 
hence  the  stress  in  it  is  equal  to  P3. 

Members  M5-L5  and  M5-L4*  The  stress  in  hanger  M5-L5  is  equal 
to  W5.  The  vertical  component  of  the  stress  in  M5-L4  is  equal  to 
^(P5  +  JF5).  Then  the  stress  in  M5-L4  is  equal  to  |(P5  +  JF5)sec<9". 

Member  U4-L4.  Considering  the  part  of  the  structure  to  the  left 
of  section  5-5  and  resolving  the  stress  in  £74-  £73-  £72  and  M5-L4  vertically 
and  horizontally  and  summing  up  the  forces  and  components,  we  obtain 


for  the  stress  in  post  £74-L4. 

Member  U4-U5-U6.  Considering  the  part  of  the  structure  to  the 
left  of  section  7-7  and  taking  moments  about  Z/6  we  can  obtain  H2,  the 
horizontal  component  of  the  stress  in  top  chord  £74-  £75-  £76.  Then  the 
stress  in  the  member  is  equal  to  H2  x  sec<£". 

Member  L4-L5-L6.  Considering  again  the  part  of  the  structure 
to  the  left  of  section  7-7  and  taking  moments  about  £74,  we  obtain 


for  the  stress  in  bottom  chord  L4-L5-L6. 

Member  M5-L6.  Considering  again  the  part  of  the  structure  to 
the  left  of  section  7-7  and  letting  F6  represent  the  vertical  component  of 
the  stress  in  top  chord  £74-£75-£76,  we  obtain 


for  the  vertical  component  of  the  stress  in  diagonal  M5-L6.  Then  mul- 
tiplying VI  by  sec#"  the  stress  in  the  member  is  obtained. 

Members  M4-M5  and  M5-M6.  These  members  have  no  direct  stress 
in  them.  They  are  for  the  purpose  of  stiffening  the  posts  £74-L4  and 
C76-L6. 

Member  U4-M5.  Considering  the  part  of  the  structure  to  the  left 
of  section  6-6  and  letting  ^(P5  +  ?F5)  and  FQ  represent  the  vertical  com- 
ponent of  the  stress  in  M5-L4  and  £74-  £75-  £76,  respectively,  and  summing 
up  the  vertical  forces  and  components  to  the  left  of  section  6-6,  we 
obtain 

F8  =  R  -  (Wl  +  W'2  .  .  .  JF4)  -  (PI  +  P2  .  .  .P4)  -  J(P5  +  7F5)  -  F6 

for  the  vertical  component  of  the  stress  in  diagonal  £74-M5.  Then  the 
stress  in  the  member  is  equal  to  F8  x  sec#". 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  485 

Member  U5-M5.  This  member  supports  the  load  Po  and  hence 
*hr?  stress  in  it  is  equal  to  Po. 

Members  U7-M7  and  M7-L7.  As  is  evident,  the  stress  in  U7-M7  is 
equal  to  PI  and  the  stress  in  M7-L7  is  equal  to  Wl. 

Members  U6-M7  and  M7-L6.  The  dead-load  stress  in  these  mem- 
bers is  due  to  the  loads  W7  and  P7.  We  will  assume  that  each  is  equally 
stressed.  Then  the  stress  in  each  will  be  equal  to  J(?F7  +  P7)sec0'". 

Member  U6-L6.  Considering  the  part  of  the  structure  to  the  left  of 
section  8-8  and  letting  VQ  and  F9  represent  the  vertical  component  of  the 
stirss  in  members  U4-U5-U6  and  M7-L6,  respectively,  and  summing  up 
ali  the  forces  and  components  to  the  left  of  section  8-8,  we  obtain 

£6  =  R  -  (Wl  +  JV2  .  .  .  JF6)  -  (PI  +P2  .  .  .P5)  -F6-F9 

for  the  stress  in  C76-L6. 

Members  U6-U7  and  L6-L7.  Considering  the  part  of  the  structure 
to  the  left  of  section  9-9  and  taking  moments  about  LI  ,  we  obtain 


for  the  stress  in  top  chord  U6-U7.  It  will  be  seen  readily  that  the 
moments  of  the  stresses  in  U6-M7  and  M7-L6  about  L7  are  equal  and  of 
opposite  signs  and  annul  each  '  other,  and  consequently  are  not  given 
in  the  equation  of  moments. 

The  stress  in  L6-L7  is  equal  to  the  stress  in  top  chord  C76-C77.  How- 
ever, the  stress  in  Z/6-L7  can  be  obtained  by  taking  moments  about  £77. 

As  the  truss  is  symmetrical  in  reference  to  the  center  of  span  and 
symmetrically  loaded,  we  have  now  sufficiently  considered  the  analytical 
determination  of  the  dead-load  stresses  in  the  truss. 

As  a  matter  of  fact,  the  dead-load  stresses  in  such  trusses  are 
graphically  determined.  The  diagram  of  the  dead-load  stresses  for  the 
right  half  of  the  truss  is  shown  at  (e).  This  diagram  is  obtained  by 
beginning  at  N  and  passing  around  the  joints  clock-  wise.  Thus,  begin- 
ning at  point  N,  we  draw  1-2  equal  to  R  and  then  1-3  and  2-3  parallel, 
respectively,  to  bN  and  BN.  Then  passing  on  to  joint  B  we  have  the 
diagram  3-2-4-5-3  for  that  joint.  Then  passing  on  to  joint  b  we  have  the 
diagram  6-1-3-5-7-6  for  that  joint.  Now,  at  C  there  are  three  unknown 
forces,  the  same  is  true  of  joint  c  and,  consequently,  before  we  can  go 
farther  we  must  determine  one  of  the  unknown  forces  at  one  of  these 
joints.  By  considering  E-M3-C  as  a  separate  truss,  as  shown  at  (f),  and 
constructing  the  diagram  at  (g),  we  have  the  stress  in  M3-C  given  by 
the  line  vu.  Then  considering  joint  C  (at  (a))  and  passing  around  the 
joint  clock-  wise,  we  have  the  part  (starting  at  7)  7-5-4-8  of  the  stress 
diagram  for  that  joint.  We  know  that  the  stress  in  post  cC  will  be  rep- 
resented by  a  vertical  line  from  7.  So  we  can  draw  7-y.  We  can  also 
draw  8-,r.  Then  laying  off  0-10  equal  to  !(P3  +  ^3)  and  drawing  9-10 
parallel  to  M3-C,  we  have  the  stress  in  M3-C  represented  by  this  line 
9-10  (see  diagram  at  (g))  and  thus  we  have  the  complete  stress  diagram 
7-5-4-8-9-10-7  for  joint  C.  For  joint  c  we  obtain  the  stress  diagram 


486 


STRUCTURAL  ENGINEERING 


(beginning  at  11)  11-6-7-10-12-11.  For  joint  d  we  obtain  the  stress 
diagram  (beginning  at  13)  13-11-12-14-13.  For  joint  D  we  obtain  the 
diagram  9-8-15-16-9.  For  joint  M3  we  obtain  the  stress  diagram  14-12- 
10-9-16-14.  The  line  16-14,  which  represents  the  stress  in  member 
A/3-2?,  must  be  parallel  to  12-10,,  which  represents  the  stress  in  member 
c-M3.  This  is  the  first  check  on  the  work. 

By  considering  G-M5-E  as  a  separate  truss,  the  stress  in  M5-E  can 
be  determined.  Now,,  considering  joint  E  and  passing  around  clock-wise, 
we  obtain  the  part  14-16-15-17  of  the  stress  diagram  for  joint  E. 

Then  drawing  14-r  and  laying  off  r-19  equal  to  ^(P5  +  W5)  and 
drawing  19-18  parallel  to  the  member  M5-E,  we  have  the  complete 


Fig.  340 


stress  diagram  14-16-15-17-18-19-14  for  joint  E.  For  joint  e  we 
obtain  the  stress  diagram  20-13-14-19-21-20.  For  joint  /  we  obtain 
the  stress  diagram  22-20-21-23-22.  For  joint  F  we  obtain  the  diagram 
18-17-24-25-18.  For  joint  M5  we  obtain  the  diagram  23-21-19-18-25-23. 
The  line  23-25  must  be  parallel  to  line  21-19.  This  is  the  second  check 
^on  the  work. 

The  dead-load  stress  is  assumed  to  be  the  same  in  the  four  members, 
M7-t/6,  M7-L6,  M7-g  and  M7-G.  This  stress  is  due  to  the  loads  Wl 
and  PI.  Then,  by  considering  LQ-M 7-G  as  a  separate  truss,  having  a 
reaction  at  each  end  of  }(W7  +  P7),  the  stress  in  M7-G  can  be  graphically 
determined. 

Starting  with  23,  we  have  the  part  23-25-24-26  of  the  stress  diagram 
for  joint  G.  Then  drawing  a  vertical  line  from  23  and  a  horizontal 
line  from  26  and  laying  off  t-28  equal  to  J(7F7  +  P7)  and  drawing  28-27 
parallel  to  M7-G,  we  obtain  the  complete  stress  diagram  23-25-24-26-27- 


DESIGN  OF  SIMPLE  EAILEOAt)  BKIDGES  487 

28-23  for  joint  G.  For  joint  g  we  have  the  stress  diagram  29-22- 
23-28-30-29. 

This  completes  the  graphical  determination  of  the  dead-load  stresses, 
as  the  truss  is  symmetrical  in  reference  to  the  center  of  span  and  the 
load  is  symmetrical  in  reference  to  same.  The  distance  26-29  should 
equal  J(P7  +  W7).  This  is  the  third  and  final  check  on  the  work. 

207.    Determination  of  Live-Load  Stresses  in  Pettit  Trusses. — 

Let  it  be  required  to  determine  the  live-load  stresses  in  the  truss  shown 
at  (a),  Fig.  340,  due  to  Cooper's  E50  loading. 

Members  bA,  AB  and  EC.  The  maximum  live-load  stress  will  occur 
in  these  members  when  the  wheel  loads  are  placed  for  maximum  shear  in 
panel  AB.  The  placing  of  the  loading  will  be  in  accordance  with  (5), 
Art.  90.  After  the  loads  are  thus  placed,  the  next  thing  to  do  is  to  deter- 
mine the  reaction  at  A  by  taking  moments  about  N.  Let  R  represent  this 
reaction.  The  next  thing  to  do  is  to  determine  the  concentration  at  A 
due  to  the  loads  in  panel  AB,  which  can  be  done  by  taking  moments  about 
B.  Let  r  represent  this  concentration.  Then  we  have  R-r  for  the  shear 
in  panel  AB.  Then  we  obtain  (R-r)secO  for  the  maximum  live-load  stress 
in  end  post  bA  and  (R-r)tanO  for  the  maximum  live-load  stress  in  each 
of  the  members  AB  and  BC  (see  Art.  174). 

Member  bC.  The  maximum  live-load  stress  will  occur  in  diagonal 
bC  when  the  loads  are  placed  according  to  the  requirements  of  (4)  of 
Art.  190.  After  the  loads  are  thus  placed,  the  next  thing  to  do  is  to 
determine  the  reaction  at  A,  which  can  be  done  by  taking  moments  about 
N.  Let  R  represent  this  reaction.  Next,  by  taking  moments  about  C, 
the  horizontal  component  of  the  stress  in  top  chord  be  is  readily  obtained. 
Let  H  represent  this  component.  Then  the  vertical  component  of  the 
stress  in  be  is  equal  to  Htan(f>.  Taking  moments  about  C  of  the  loads 
in  panel  BC,  the  concentration  at  B  is  readily  obtained.  Let  r  repre- 
sent this  concentration.  Now,  for  the  vertical  component  of  the  stress 
in  diagonal  bC,  we  have  V  =  R  -  r  -  J/tan<£,  and  multiplying  this  by  sec0, 
we  obtain  (R  —  r  —  Htan^secO  for  the  stress  in  the  member  bC. 

Member  bB.  The  maximum  live-load  stress  in  this  hanger,  as  is 
obvious,  is  equal  to  the  maximum  live-load  floor  beam  concentration,  which 
is  determined  as  explained  in  Art.  148. 

Member  be.  The  maximum  live-load  stress  will  occur  in  this  member 
when  the  loads  are  placed  for  maximum  moment  about  C.  The  placing 
of  the  loads  will  be  in  accordance  with  (5),  Art.  91.  That  is,  a  wheel 
must  be  at  C  and  the  average  unit  load  to  the  left  of  C  must  equal 
(approximately)  the  unit  load  on  the  span.  After  the  loads  are  thus  placed 
the  next  thing  to  do  is  to  determine  the  reaction  at  A,  which  can  be  done 
by  taking  moments  about  N.  Then,  by  taking  moments  about  C  of  the 
forces  to  the  left,  the  horizontal  component  of  the  stress  in  be  is  readily 
obtained,  and  multiplying  this  component  by  sec<£,  the  maximum  live-load 
stress  in  be  is  obtained. 

Members  cC  and  c-Ml.  It  is  obvious  if  C-M\-E  be  considered  a 
separate  truss,  shown  as  C-o-E  at  (b),  that  the  stress  in  cC  and  c-M\  will 
not  be  affected  in  the  least.  The  truss  C-o-E  really  acts  as  a  stringer 
extending  from  C  to  E.  Then,  as  is  readily  seen,  the  maximum  live-load 
stress  in  cC  can  be  obtained  by  loading  the  span  from  the  right,  placing 


488  STRUCTURAL-  ENGINEERING 

a  wheel  at  E  and  loading  panel  CE  (no  loads  to  the  left  of  C),  all  in 
accordance  with  (4)  of  Art.  190. 

In  applying  equation  (4)  s  should  be  taken  equal  to  the  distance 
I  A,  shown  at  (b),  and  a  should  be  taken  equal  to  distance  AC. 

After  the  loading  is  placed,,  the  next  thing  to  do  is  to  determine  the 
reaction  at  A,  which  can  be  done  by  taking  moments  about  N.  Let  R 
represent  this  reaction.  The  next  thing  to  do  is  to  determine  the  con- 
centration at  C  due  to  the  loads  in  panel  CE,  which  can  be  done  by 
taking  moments  about  E,  ignoring  panel  -point  D  altogether ;  that  is,  CE 
would  be  considered  as  a  stringer.  Let  r  represent  this  concentration 
atC. 

Next,  by  taking  moments  about  C,  the  horizontal  component  of  the 
stress  in  top  chord  be  is  readily  obtained,  as  previously  explained.  Let 
H  represent  this  component.  Then  the  vertical  component  of  the  stress 
in  the  top  chord  be  is  equal  to  HtaiKf>.  Now,  considering  the  part  of  the 
structure  to  the  left  of  section  1-1,  and  adding  up  the  vertical  components 
and  forces,  we  obtain  R-r-Htan<f>  for  the  maximum  live-load  stress  in 
member  cC.  The  member  oC  is  entirely  ignored,  as  the  vertical  com- 
ponent of  its  stress  is  included  in  the  concentration  r. 

To  obtain  the  maximum  live-load  stress  in  c-Ml,  the  loads  would  be 
placed  very  much  the  same  as  for  cC.  In  applying  equation  (4)  of  Art. 
190  *  would  be  taken  equal  to  the  distance  I' A  and  a  equal  to  AC.  After 
the  loads  are  properly  placed,  the  next  thing  to  do  is  to  determine  the 
reaction  at  A,  which  can  be  done  by  taking  moments  about  N.  Then 
the  next  thing  to  do  is  to  determine  the  concentration  at  C  due  to  the 
loads  in  panel  CE,  ignoring  panel  D.  By  taking  moments  about  E,  the 
horizontal  component  of  the  stress  in  top  chord  ce  is  readily  obtained,  and 
multiplying  this  component  by  tan</>',  the  vertical  component  of  the  stress 
in  top  chord  ce  is  obtained.  Then  considering  the  part  of  the  structure  to 
the  left  of  section  2-2  and  summing  up  the  vertical  components  and  forces 
to  the  left  which  consist  of  the  reaction  at  A,  concentration  at  C,  the 
vertical  component  of  the  stress  in  ce,  and  the  vertical  component  of  the 
stress  in  cMl,  the  vertical  component  of  the  stress  in  cMl  cstn  be 
obtained;  and  multiplying  this  component  by  sec0',  the  stress  in  cM\ 
is  obtained. 

Members  D-M1  and  C-M1.  As  is  obvious,  the  maximum  live-load 
stress  in  hanger  D-M 1  is  equal  to  the  maximum  live-load  floor  beam  concen- 
tration, which  is  determined  as  explained  in  Art.  148.  The  maximum  live- 
load  stress  in  sub-diagonal  C-M1,  as  is  readily  seen,  is  equal  to  one-half 
of  the  maximum  floor  beam  concentration  multiplied  by  sec#'. 

Member  ce.  The  maximum  live-load  stress  in  this  member  occurs 
when  a  load  is  at  E  and  the  average  unit  load  to  the  left  of  E  is  equal 
to  the  average  unit  load  on  the  span.  This  is  in  accordance  with  Art. 
91.  After  the  loads  are  properly  placed  the  next  thing  to  do  is  to 
determine  the  reaction  at  A,  which  can  be  done  by  taking  moments  about 
N.  Then  considering  the  part  of  the  structure  to  the  left  of  section  3-3 
and  taking  moments  about  E,  the  horizontal  component  of  the  stress  in 
top  chord  ce  is  readily  obtained,  and  then  multiplying  this  component 
by  sec<£',  the  maximum  live-load  stress  in  ce  is  obtained. 

Member  E-M1.  It  is  readily  seen  that  if  oE  (shown  at  (b))  were 
combined  with  E-M1,  any  load  at  D  would  cause  compression  in  E-M1 


DESIGN  OF  SIMPLE  EAILEOAD  BEIDGES  489 

and  hence  the  member  E-M\,  as  far  as  maximum  stress  is  concerned, 
acts  just  the  same  as  an  ordinary  diagonal  where  DE  is  the  panel 
length.  Then  the  maximum  stress  in  E-M1  is  obtained  by  loading  the 
span  from  the  right  with  a  wheel  at  Ef  and  loading  panel  DE  (no  loads 
to  the  left  of  D)  according  to  (4)  of  Art.  190.  In  applying  equation 
(4)  s  should  be  taken  equal  to  the  distance  I' A  and  a  equal  to  the  dis- 
tance AD.  After  the  loads  are  properly  placed,  the  next  thing  to  do  is 
to  determine  the  reaction  at  A,  which  can  be  done  by  taking  moments 
about  N.  Then  the  next  thing  to  do  is  to  determine  the  concentration 
at  D,  due  to  the  loads  in  panel  DE.  Then,  considering  the  forces  to  the 
left  of  section  3-3  (oE  and  E-M\  being  considered  as  combined,  that  is, 
one  member)  and  taking  moments  about  E,  the  horizontal  component  of 
the  stress  in  top  chord  ce  is  readily  obtained;  and  multiplying  this 
horizontal  component  by  tanc//  the  vertical  component  of  the  stress  in  the 
top  chord  ce  is  obtained.  Then  summing  up  the  vertical  components  and 
forces  to  the  left  of  section  3-3  the  vertical  component  of  the  stress  in 
diagonal  E-Ml  is  readily  determined ;  and  multiplying  this  component  by 
secQ'  the  maximum  live-load  stress  E-Ml  is  obtained. 

Member  CE.  It  is  readily  seen  that  any  load  in  panel  CD  or  DE 
will  cause  tension  in  CE,  owing  to  the  member  CE  acting  as  the  bottom 
chord  of  truss  C-Ml-E.  This  stress,  as  we  may  say,  is  in  addition  to 
the  stress  produced  in  the  member  acting  as  a  main  bottom  chord  sec- 
tion of  the  structure.  Then  it  appears  that  the  maximum  live-load  stress 
in  bottom  chord  CE  will  occur  when  the  loads  in  panels  CD  and  DE 
have  some  certain  value  as  compared  to  the  other  loads  on  the  structure. 
This  can  be  investigated  most  satisfactorily  by  the  use  of  influence  lines. 

The  stress  in  CE  is  finally  obtained  by  considering  the  part  of  the 
structure  to  the  left  of  section  3-3  and  taking  moments  about  c.  Then 
evidently  the  stress  in  CE  will  vary  as  the  moments  about  c  and  hence 
the  stress  in  the  member  can  be  investigated  by  constructing  an  influence 
line  for  moments  about  c. 

Let  P  represent  a  load  at  any  point  x  distance  from  N.  Then  when 
P  is  at  any  point  to  the  right  of  panel  point  E  we  have  M—(Px/L~)a 
for  the  bending  moment  about  c.  If  x=b,  the  last  equation  becomes 
M=(Pb/L)a,  and  if  P  =  l  we  have  M=(b/L)a.  So  if  we  draw  00' 
at  (c)  and  lay  off  uv  equal  to  (b/L)a  we  can  draw  uO' ',  which  is  the 
influence  line  for  moments  about  c  for  loads  to  the  right  of  E. 

If  the  load  P  moves  to  any  point  in  panel  DE,  the  moment 
about  c  is 


If    x-b    we    have    the    same    as    given    above    for    ordinate    uv.      If 
and  P  =  l  the  equation  reduces  to 


,     a  +  d 
M  =  -L-    a  +  d  =  ~L 

Then  if  the  ordinate  nm  be  laid  off  equal  to  this  last  value  of  M'  the 
line  nu  can  be  drawn,  which  is  the  influence  line  for  moments  about 
c  for  loads  in  panel  DE. 


490  STRUCTURAL  ENGINEERING 

If  the  load  P  moves  to  any  point  in  panel  CD,  the  moment  about  c 
will  be 


This  is  the  equation  to  the  line  in,  which  is  the  influence  line  for  moments 
about  c  for  loads  in  panel  CD. 

If  the  load  P  moves  to  any  point  to  the  left  of  C,  the  moment  about 
c  will  be 

M"'  =  (Pj\  a-(x-b-2d)P  = 

which  is  the  equation  to  line  tO.  But  this  value  of  M"'  is  exactly  the 
same  as  found  above  for  M".  Therefore,  the  lines  tn  and  tO  are  in  the 
same  straight  line  On  and  hence  On  is  the  influence  line  for  moments 
about  c  for  loads  to  the  left  of  D.  Then  evidently  OnuO'  is  the  complete 
influence  line  for  moments  about  c.  As  is  seen,  this  influence  line  has 
three  different  slopes  and  consequently  there  will  be  three  different 
moment  increments  to  consider. 

Let  Wl  represent  the  resultant  of  the  loads  to  the  left  of  D,  W2 
the  resultant  of  the  loads  in  panel  DE,  W3  the  resultant  of  the  loads  to 
the  right  of  E  and  let  W  represent  the  total  load  on  the  span;  that  is, 
W=W1  +  W2  +  W3.  Then  for  the  moment  about  c,  we  have 


(1). 

Now  if  all  loads  move  to  the  left  the  distance  dx,  we  have 

AM  =  -  Wldxtanfil  +  W2dxianp2  +  W3dxtanf33    (2) . 

for  the  increment  of  the  moment  about  c.  If  the  loads  are  in  the  posi- 
tion for  maximum  moment  about  c  this  increment  will  be  equal  to  0  and, 
hence,  (2)  would  become 

0  =  -  Wltanpl  +  W2t&n/32  +  JF3tan£3    (3). 

But,  tan/?l  =  —f^' 

L  +  a 


la  + 
=    - 


tan/33  =  -  • 
Li 

Substituting  these  values  of  the  tangents  in  (3),  we  obtain 

a 
L' 

from  which  we  obtain 


(W2  -  Wl)L  +  a(Wl  +  W2  +  7P3)  =0, 
from  which  we  obtain 

W  _W1-W2 
L  =          a 


DESIGN  OF  SIMPLE  EAILEOAD  BKIDGES  491 

Expressing  equation  (4)  in  words,  the  average  unit  load  on  the  span 
is  equal  to  the  load  to  the  left  of  D,  minus  the  load  in  panel  DE,  divided 
by  a.  This  can  be  taken  as  the  criterion  for  placing  the  load  for  maxi- 
mum stress  in  bottom  chord  CE. 

After  the  loads  are  placed  in  accordance  with  equation  (4),  the  next 
thing  to  do  is  to  determine  the  reaction  at  A,  which  can  be  done  by  taking 
moments  about  N.  Then  the  next  thing  is  to  determine  the  concentra- 
tion at  D,  which  can  be  done  by  taking  moments  about  C  and  E. 

Let  R  —  the  reaction  at  A,  r  =  the  concentration  at  D  (due  to  the 
loads  in  panels  CD  and  DE),  m  =  moment  of  load  to  the  left  of  C  about 
c,  and  h  =  height  of  post  cC.  Then  considering  the  part  of  the  structure 
to  the  left  of  section  5-3  and  taking  moments  about  c,  we  obtain 

S=(Ra-m  +  rd}- 
n 

for  the  maximum  live-load  stress  in  bottom  chord  CE. 

The  moment  about  c  of  course  could  be  obtained  by  the  use  of  influ- 
ence line  OnuO' ' . 

If  W  remains  constant,  the  increment  of  the  moment  would  change 
signs  only  as  a  load  passed  joint  D,  so  there  will  be  a  load  at  that  joint 
when  the  maximum  moment  occurs  at  joint  c.  Then  in  placing  the  load- 
ing for  maximum  moment  about  c,  that  is,  satisfying  (4),  a  wheel  must 
be  at  joint  D. 

Member  eg.  The  maximum  live-load  stress  in  top  chord  eg  is 
obtained  by  placing  a  load  at  G  such  that  the  unit  load  to  the  left  of  G 
is  equal  to  the  unit  load  on  the  span.  Then  by  taking  moments  about  G 
the  horizontal  component  of  the  stress  in  eg  is  readily  determined  and 
multiplying  this  component  by  sec<£"  the  stress  in  the  member  is  obtained. 

Member  eE.  The  maximum  live-load  stress  in  this  member  is 
determined  in  the  same  manner  as  shown  above  for  cC.  The  sub-diag- 
onal E-M3  and  hanger  F-M3  are  assumed  to  be  omitted.  The  span  is 
loaded  from  the  right,  a  wheel  at  G  and  panel  EG  loaded  in  accordance 
with  (4)  of  Art.  190. 

The  value  of  s  (in  equation  (4))  is  obtained  by  prolonging  top  chord 
ce  and  a  is  taken  equal  to  the  distance  AE.  After  the  loads  are  prop- 
erly placed,  the  next  thing  to  do  is  to  determine  the  reaction  at  A,  which 
can  be  done  by  taking  moments  about  N.  Next,  the  concentration  at  E 
can  be  determined  by  taking  moments  about  G.  Then,  the  next  thing 
is  to  determine  the  vertical  component  in  top  chord  ce,  which  can  be  done 
by  first  taking  moments  about  E,  thereby  obtaining  the  horizontal  com- 
ponent of  the  stress  in  the  member  and  multiplying  this  component  by 
tan<£',  we  obtain  the  vertical  component  of  the  stress  in  ce. 

Let  R  represent  the  reaction  at  A,  r  the  concentration  at  E,  and  V 
the  vertical  component  of  the  stress  in  top  chord  ce.  Then  considering 
the  part  of  the  structure  to  the  left  of  section  4-4  and  summing  up  the 
vertical  components  and  forces,  we  have 

S  =  R-r-F, 

for  the  maximum  live-load  stress  in  post  eE. 

Member  e-M3.  The  maximum  live-load  stress  in  this  member  is 
determined  in  the  same  manner  as  shown  above  for  member  c-Ml.  The 


492  STEUCTUEAL  ENGINEEEING 

members  E-M3  and  F-M3  are  assumed  to  be  omitted.  The  span  is  loaded 
from  the  right  with  a  wheel  at  G  and  panel  EG  loaded  in  accordance 
with  (4)  of  Art.  190.  Top  chord  eg  in  that  case  would  be  prolonged  to 
determine  the  value  of  s,  and  a  would  be  the  distance  AE. 

After  the  loads  are  properly  placed  for  maximum  stress  in  e-M3,  the 
next  thing  to  do  is  to  determine  the  reaction  at  A,  which  can  be  done  by 
taking  moments  about  N.  The  next  thing  to  do  is  to  determine  the  con- 
centration at  E,  which  can  be  done  by  taking  moments  about  G.  Then 
the  next  thing  in  order  is  to  determine  the  vertical  component  of  the 
stress  in  top  chord  eg,  which  can  be  done  by  first  taking  moments  about 
G,  thus  obtaining  the  horizontal  component  of  the  stress  in  top  chord  eg 
and  multiplying  this  component  by  tan</>",  we  obtain  the  vertical  com- 
ponent of  top  chord  eg.  Then  considering  the  part  of  the  structure  to 
the  left  of  section  5-5,  and  summing  up  the  vertical  components  and 
forces,  we  obtain 

Vl=R-r-V 

for  the  vertical  component  of  the  stress  in  diagonal  e-M3,  and  multiply- 
ing this  component  by  sec0",  we  obtain  the  maximum  live-load  stress  in 
member  e-M3. 

Members  F-M3  and  E-M3.  As  is  evident,  the  maximum  live-load 
stress  in  hanger  F-M3  is  equal  to  the  maximum  floor  beam  concentration, 
which  is  determined  as  explained  in  Art.  148.  The  maximum  live-load 
stress  in  sub-diagonal  E-M3  is  equal  to  one-half  of  this  concentration 
multiplied  by  sec0". 

Member  G-M3.  The  maximum  live-load  stress  in  this  member  is 
determined  in  the  same  manner  as  explained  above  for  diagonal  E-M1. 
The  span  is  loaded  from  the  right,  a  wheel  is  placed  at  G  and  panel 
GF  is  loaded  in  accordance  with  (4)  of  Art.  190.  In  this  case  the  value  s 
is  obtained  by  prolonging  top  chord  eg  (see  Art.  190)  and  a  is  equal  to 
the  distance  AF.  The  reaction  at  A  is  obtained  by  taking  moments  about 
N.  The  concentration  at  F  is  obtained  by  taking  moments  about  G,  and 
the  horizontal  component  of  the  stress  in  top  chord  eg  is  obtained  by 
taking  moments  about  G,  and  the  vertical  component  of  same  is  then 
obtained  by  multiplying  the  horizontal  component  by  tan</>".  After  these 
are  determined  the  vertical  component  of  the  stress  in  diagonal  G-M3  is 
obtained  by  considering  the  part  of  the  structure  to  the  left  of  section  6-6 
and  summing  up  the  vertical  component  and  forces,  and  then  the  stress 
in  G-M3  is  obtained  by  multiplying  this  component  by  sec0". 

Member  EG.  The  maximum  live-load  stress  in  this  member  is  deter- 
mined in  the  same  manner  as  explained  above  for  bottom  chord  CE.  In 
applying  equation  (4)  (given  above)  a  would  be  taken  equal  to  distance 
AE.  W\  would  be  the  load  between  A  and  F,  W2  the  load  in  panel  FG, 
W3  the  load  between  G  and  Nt  and  W  the  total  load  on  the  span. 

After  the  loads  are  properly  placed,  that  is,  in  accordance  with 
equation  (4),  the  stress  in  EG  is  readily  obtained  by  considering  the 
part  of  the  structure  to  the  left  of  section  6-6  and  taking  moments 
about  e. 

Member  gG.  The  maximum  live-load  stress  in  this  member  is  ob- 
tained by  considering  members  Gk  and  Hh  as  being  omitted.  The 
span  is  then  loaded  from  the  right,  a  wheel  at  K  and  panel  KG  loaded 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  493 

in  accordance  with  (4)  of  Art.  190.  The  value  of  s  in  equation  (4)  of 
Art.  190  is  obtained  by  prolonging  top  chord  eg  and  a  is  equal  to  the  dis- 
tance AG.  After  the  loads  are  properly  placed  the  reaction  at  A  is 
obtained  by  taking  moments  about  N.  The  concentration  at  G  is  obtained 
by  taking  moments  about  K.  The  horizontal  component  of  the  stress  in 
top  chord  eg  is  obtained  by  taking  moments  about  G  and  the  vertical 
component  of  same  is  obtained  by  multiplying  the  horizontal  component 
by  tan<£".  Having  the  above  determined.,  the  stress  in  gG  is  obtained 
by  summing  up  the  vertical  forces  to  the  left  of  section  7-7  (ignoring 
member  G-M5). 

Member  GK.  The  maximum  live-load  stress  in  this  member  is  de- 
termined in  the  same  manner  as  explained  above  for  bottom  chord  CE. 
In  applying  equation  (4)  (given  above)  a  would  be  taken  equal  to  dis- 
tance AG,  WL  would  be  the  load  between  A  and  H,  W2  the  load  in  panel 
HK,  W3  the  load  between  K  and  2V.  After  the  loads  are  properly  placed, 
that  is,,  in  accordance  with  equation  (4),  the  stress  in  GK  is  readily  ob- 
tained by  considering  the  part  of  the  structure  to  the-  left  of  section  8-8 
(ignoring  member  Ar-M5)  and  taking  moments  about  g. 

Member  g-Mo.  To  obtain  the  maximum  live-load  stress  in  this 
member,  we  assume  the  members  Gk  and  Hh  to  be  omitted,  and  load 
the  span  according  to  (5)  of  Art.  90.  That  is,  the  span  would  be  loaded 
from  the  right  with  a  wheel  at  K  arid  the  load  in  panel  GK  equal  to  the 
load  on  the  bridge  divided  by  the  number  of  panels.  The  panels  con- 
sidered in  this  case  would  be  double  panels  —  that  is,  the  number  would 
be  7  instead  of  14. 

After  having  the  loads  thus  placed,  the  next  thing  to  do  is  to  deter- 
mine the  reaction  at  A  and  the  concentration  at  G.  These  can  be  ob- 
tained by  taking  moments  about  2V  and  K,  respectively.  Let  R  repre- 
sent the  reaction  at  A  and  r  the  concentration  at  G.  Then  we  have 


for  the  maximum  live-load  stress  in  diagonal  g-M5. 

Member  K-M5.  To  obtain  the  maximum  live-load  stress  in  this 
member  we  assume  member  k-M5  to  be  omitted  and  load  the  span  accord- 
ing to  (5)  of  Art.  90.  That  is,  the  span  would  be  loaded  from  the  right 
with  a  wheel  at  K,  and  the  load  in  panel  HK  equal  to  the  total  load  on 
the  span  divided  by  the  number  of  panels.  In  this  case  the  number  of 
panels  would  be  14.  After  the  loads  are  thus  placed,  the  next  thing 
to  do  is  to  determine  the  reaction  at  A  and  the  concentration  at  H.  Let 
R'  represent  the  reaction  at  A,  and  r'  the  concentration  at  H.  Then  we 
have 

S'=(R'-r')secOf" 

for  the  maximum  live-load  stress  in  diagonal  K-M5. 

Members  g-M5  and  k-M5  would  be  assumed  to  have  equal  tensile 
stress  and  likewise  members  K-M5  and  G-M5.  The  latter  members  would 
be  subjected  to  compression  from  the  floor  beam  concentration  at  H. 
Assuming  the  concentration  at  H  to  be  transmitted  to  panel  points  G  and 
K  by  sub-truss  K-M5-G  the  maximum  live-load  compression  in  each  of 
the  members  G-M5  and  K-M5  would  be  equal  to  one-half  of  the  maximum 
floor  beam  concentration  at  H  multiplied  by  sec#"'. 


494 


STRUCTURAL  ENGINEERING 


Member  gk.  The  maximum  live-load  stress  in  this  member  is  ob- 
tained by  placing  the  load  for  maximum  moments  about  K.  Then  taking 
moments  about  K  (ignoring  member  &-M5)  and  dividing  this  moment  by 
the  height  of  post  kK,  we  would  obtain  the  greater  part  of  the  maximum 
stress  in  gk.  Next  the  concentration  at  H,  due  to  the  loads  in  panels 
GH  and  HK,  can  be  determined  by  taking  moments  about  G  and  K. 
Then,  multiplying  one-fourth  of  this  concentration  at  H  by  tan#"',  and 
adding  the  result  to  the  stress  found  in  gk,  by  taking  moments  about  K, 
the  total  maximum  live-load  stress  in  the  member  gk  is  obtained. 

Maximum  Tension  in  Post  gG.  The  maximum  live-load  tension  will 
occur  in  this  post  when  the  live  load  extends  from  A  to  some  point  beyond 
G,  so  that  the  stress  in  diagonal  g-M5  is  zero.  The  position  of  the  load 


Fig.  341 

can  be  determined  by  trial  as  explained  in  Art.  205  for  curved  chord 
Pratt  trusses.  After  the  position  of  the  loading  is  found,  the  tension 
in  the  member  is  readily  determined  by  considering  the  part  of  the  struc- 
ture to  the  left  of  section  7-7.  The  use  of  an  equivalent  uniform  live- 
load  will  simplify  the  work  very  much  and  the  result  thus  obtained 
will  be  sufficiently  accurate. 

Stress  in  Counters.  In  case  a  member  g-M3  were  inserted,  it  would 
be  known  as  a  counter.  This  member  would  not  be  stressed  at  all 
unless  the  dead-load  tension  in  either  or  both  of  the  members  e--M3  and 
G-M3  were  reversed.  In  case  either,  or  both,  of  these  members  be  sub- 
jected to  reversal,  the  counter  g-M3  would  be  used  or  members  e-M3  and 
G-M3  would  be  designed  to  carry  both  tension  and  compression.  In 


DESIGN  OF  SIMPLE  KAILROAD  BRIDGES  495 

case  the  counter  g-M3  is  required,  the  maximum  live-load  stress  (tension) 
in  it  can  be  determined  by  assuming  members  eG  and  F-M3  to  be  omitted 
and  loading  the  span  from  the  left  with  a  load  at  E  and  loading  panel 
EG  in  accordance  with  (4)  of  Art.  190.  After  the  loading  is  thus  placed, 
the  next  thing  to  do  is  to  determine  the  reaction  at  N  and  the  concentra- 
tion at  G,  which  can  be  done  by  taking  moments  about  A  and  E,  re- 
spectively. Then  by  considering  the  part  of  the  structure  to  the  right 
of  section  6-6,  the  stress  is  readily  determined.  If  counters  are  found 
necessary  at  other  points  the  stress  in  them  can  be  determined  in  a 
similar  manner. 

The  determination  of  the  stresses  in  trusses  having  the  sub-paneling 
at  the  top  of  the  truss,  as  shown  at  (a)  in  Fig.  341,  is  very  similar  to 
that  given  above  for  the  truss  shown  in  Fig.  339  wherein  the  sub-paneling 
is  at  the  bottom  of  the  truss. 

Dead-Load  Stresses.  The  graphical  determination  of  the  dead-load 
stresses  in  the  type  of  truss  shown  in  Fig.  341  is  simpler  than  in  the 
case  of  the  type  shown  in  Fig.  339,  as  the  sub-paneling  does  not  bother 
at  all  in  the  case  of  the  truss  shown  in  Fig.  341. 

The  graphical  diagram  of  the  dead-load  stresses  for  the  right  half 
of  the  truss  is  shown  at  (b).  This  diagram  is  readily  followed  throughout. 

The  analytical  determination  of  the  dead-load  stresses  is  just  the 
same  in  some  cases  and  similar  in  other  cases  to  that  given  above  for  the 
truss  shown  in  Fig.  339.  To  save  space,  just  the  formulas  for  the  dead- 
load  str.esses  in  part  of  the  members  in  the  right  half  of  the  truss  shown 
in  Fig.  341  will  be  given: 

Stress  in  sT  equals  RsecO. 

Stress  in  TS  and  OS  equals  RtanO. 

Stress  in  sS  equals  W\. 

Stress  in  os  equals  [(R  x  2d)  -  (Wl  +  Pl)d]sec<f>/hl  (d  =  panel  Igth.). 

Stress  in  Os  equals  [-R  -  (Wl  +  PI)  -  F]sec#,  where  V  represents  the 
vertical  component  of  the  stress  in  top  chord  os. 

The  stress  in  member  oO  equals  R  -  (Wl  +  W2  +  P1)  -  V.  In  this 
case  the  part  of  the  structure  to  the  right  of  section  1-1  is  considered. 

The  stress  in  member  MO  equals  [Rx2d- (Wl+Pl)d]l/hl.  This 
is  obtained  by  taking  moments  about  o. 

The  stress  in  member  mo  equals  [(R  x4d)-  (3W1  +  3P1+2W2 
+  2P2)d]seccf)l/h2.  This  is  obtained  by  considering  the  part  of  the 
structure  to  the  right  of  section  2-2  and  taking  moments  about  M. 

The  stress  in  member  o-M3  equals  [R-  (W.1  +  W2  +  P1 +  P2) 
-Fl]sec#l,  where  V\  represents  the  vertical  component  of  the  stress  in 
top  chord  mo.  In  this  last  case  the  part  of  the  structure  to  the  right  of 
section  2-2  was  considered. 

The  stress  in  sub-post  n-M3  is  equal  to  P3  and  in  N-M3  it  is  W3. 

The  stress  in  sub-diagonal  m-M3  is  determined  by  considering 
n-M3-o  as  an  independent  truss  supporting  the  load  (W3+P3)  at  M3,  as 
shown  at  (c)  in  Fig.  341. 

The  stress  in  the  member  m-M3  equals  (JF3  +  P3)  -f  (cos02 
+  sin02cos<91). 

The  stress  in  member  m-M3  can  be  determined  most  readily  by 
graphics  (see  the  diagram  at  (d)). 

The  stress  in  M-M3  equals    [12-  (W1  +  W2  +  JF3 


496  STRUCTURAL  ENGINEERING 

Fl  +  F2]sec#l,  where  V\  and  JT2  represent,  respectively.,  the  vertical  com- 
ponent of  the  stress  in  mo  and  m-M 3.  In  this  case  the  part  of  the  structure 
to  the  right  of  section  3-3  is  considered. 

The  stress  in  mM  equals  R  -  (Wl  +  W2  +  W3  +  W4=  +  P1  +  P2  +  P3) 
-  V\  +  F2.  In  this  case  the  part  of  the  structure  to  the  right  of  section 
4-4  is  considered. 

Following  this  method  the  dead-load  stresses  in  the  other  members 
of  the  truss  are  readily  determined. 

Live-Load  Stresses.  The  determination  of  the  live-load  stresses  in 
the  type  of  truss  shown  in  Fig.  341  is  very  much  the  same  as  given  above 
for  the  truss  shown  in  Fig.  340.  The  method  of  analysis  will  be  suffi- 
ciently shown  by  considering  the  members  in  panel  EG. 

Member  eE.  The  maximum  live-load  stress  will  occur  in  this  mem- 
ber when  the  span  is  loaded  from  the  right,  a  wheel  at  G  and  panel  EG 
loaded  in  accordance  with  (4)  of  Art.  190.  In  this  case  the  top  chord  ce 
would  be  prolonged  to  determine  the  value  of  s  to  be  used  in  equation 
(4)  of  Art.  190  and  a  would  be  taken  equal  to  distance  AE.  After  the 
loads  are  properly  placed  for  maximum  stress  in  eE,  the  next  thing  to 
do  is  to  determine  the  reaction  at  A  and  the  concentration  at  E,  which 
can  be  done  by  taking  moments  about  T  and  G,  respectively.  The  next 
thing  to  do  is  to  determine  the  vertical  component  of  the  stress  in  top 
chord  ce.  This  can  be  done  by  taking  moments  about  E,  thus  obtaining 
the  horizontal  component  of  the  stress  in  ce  (ignoring  member  e-M3,  as 
it  is  not  subjected  to  any  live-load  stress  at  this  time,  the  live  load  not 
extending  beyond  £),  and  then  multiplying  this  horizontal  component 
by  tan<£l,  the  vertical  component  of  the  stress  in  ce  is  obtained.  Let  R 
represent  the  reaction  at  A,  r  the  concentration  at  E,  and  V  the  vertical 
component  of  the  stress  in  top  chord  ce.  Then  considering  the  part  of 
the  structure  to  the  left  of  section  5-5  and  summing  up  the  vertical  com- 
ponents and  forces,  we  have 

S=R-r-V 

for  the  maximum  live-load  stress  in  post  eE. 

Member  EG.  The  stress  in  this  member  is  obtained,  as  is  obvious, 
by  taking  moments  about  joint  e.  Then  by  placing  the  loads  so  that 
there  is  a  wheel  at  E  with  the  unit  load  to  the  left  of  E  equal  to  the  unit 
load  on  the  span,  which  is  in  accordance  with  Art.  91,  and  taking 
moments  about  e  of  the  forces  to  the  left  and  dividing  this  moment  by 
the  height  of  post  eE,  the  maximum  live-load  stress  in  EG  is  readily 
determined. 

Member  F-M5.  The  maximum  live-load  stress  in  this  hanger  is 
equal  to  the  maximum  live-load  floor  beam  concentration  at  F,  which 
is  determined  as  explained  in  Art.  148. 

Members  M4*M5,  M5-M6  and  f-M5.  These  members  are  not  sub- 
jected to  live-load  stress  at  all.  /-M5  is  subjected  to  dead-load  stress 
from  the  load  at  /  only.  The  other  two  members  are  not  subjected  to 
any  direct  stress  whatever.  They  are  for  the  purpose  of  stiffening  the 
posts  eE  and  cjG. 

Member  e-M5.  The  maximum  live-load  stress  in  this  member  will 
occur  when  the  span  is  loaded  from  the  right,  a  wheel  at  F  and  panel 
EF  loaded  in  accordance  with  (4)  of  Art.  190. 


DESIGN  OF  SIMPLE  EAILROAD  BEIDGES  497 

The  top  chord  eg  would  be  prolonged  to  determine  the  value  of  s  to 
use  in  (4)  of  Art.  190,  and  a  would  be  taken  equal  to  distance  AE. 
After  the  loads  are  properly  placed  for  maximum  live-load  stress  in  diag- 
onal e-M  5,  the  next  thing  to  do  is  to  determine  the  reaction  at  A  and  the 
concentration  at  E,  which  can  be  done  by  taking  moments  about  T  and  F, 
respectively.  Let  R  represent  the  reaction  at  A  and  r  the  concentration 
at  E.  Then  considering  the  part  of  the  structure  to  the  left  of  section 
6-6  and  summing  up  the  vertical  components  and  forces,  we  have 

S=(R-r-F)sec63, 

for  the  maximum  live-load  stress  in  diagonal  e-M5.  V  here  represents 
the  vertical  component  of  the  stress  in  top  chord  eg,  which  can  be  deter- 
mined by  considering  the  part  of  the  structure  to  the  left  of  section  6-6, 
and  taking  moments  about  G  and  dividing  this  moment  by  the  height  of 
post  gG,  thus  obtaining  the  horizontal  component  of  the  stress  in  top 
chord  eg,  then  multiplying  this  component  by  tan$2,  the  vertical  com- 
ponent V  is  obtained. 

Member  G-Mo.  In  determining  the  maximum  live-load  stress  in 
this  member,,  the  members  F-M5,  g-M5  and  f-M  5  are  assumed  to  be 
omitted.  The  maximum  stress  will  occur  when  the  span  is  loaded  from 
the  right,  a  wheel  at  G,  and  panel  GE  loaded  according  to  (4)  of  Art. 
190.  After  the  loads  are  properly  placed  for  maximum  stress  in  diagonal 
G-M5,  the  stress  is  obtained  by  considering  the  part  of  the  structure 
to  the  left  of  section  7-7  (ignoring  member  g-M5)  and  summing  up  the 
vertical  components  and  forces.  Thus  let  R  represent  the  reaction  at  A, 
r  the  concentration  at  E  and  V\  the  vertical  component  of  the  stress  in 
top  chord  eg,  then  we  have 


for  the  maximum  live-load  stress  in  diagonal  G-M5. 

Member  g-M5.  The  maximum  live-load  stress  will  occur  in  this 
member  when  the  live-load  concentration  at  F  is  a  maximum.  Let  r 
represent  this  concentration.  Then,  considering  e-M5-g  as  an  independent 
truss,  we  obtain 

S  =  r  -r  (cos04  +  sin04cos03) 

for  the  maximum  live-load  stress  in  sub-diagonal  g-M5.  This  stress  in 
g-M5  can  be  determined  most  readily  by  graphics,  considering  e-M5-g 
as  an  independent  truss. 

Member  eg.  The  live-load  stress  in  this  member  can  be  investigated 
most  satisfactorily  by  the  use  of  the  influence  line. 

The  stress  in  eg  is  finally  obtained  by  considering  the  part  of  the 
structure  to  the  left  of  section  6-6  and  taking  moments  about  G.  Then 
evidently  the  stress  in  eg  will  vary  directly  as  the  moments  about  G  and, 
hence,  the  stress  in  the  member  can  be  investigated  by  constructing  an 
influence  line  for  moments  about  G. 

Let  P  represent  a  load  at  any  point  x  distance  from  T.  Then 
when  P  is  at  any  point  to  the  right  of  panel  point  G,  we  have 


498  STRUCTURAL  ENGINEERING 

for  the  bending  moment  about  G.     If  x-\j,  the  above  equation  becomes 


and  if  P  =  l, 


So   if   we   draw    00'    at    (e)    and   lay   off   ts  —  ba/L,   we   can   draw    tO' 
which  is  the  influence  line  for  moments  about  G  for  loads  to  the  right  of  G. 
If  the  load  P  moves  to  any  point  in  panel  FG,  the  moment  about  G, 
considering  the  part  of  the  structure  to  the  left  of  section  G-6,  is 

fP4 

m  a, 


w 


which  is  the  same  as   found  above  for  M,  so  the  influence  line  vt  for 
loads  in  panel  FG  is  simply  a  continuation  of  the  line  tO'  '.     If  x  = 
and  P  =  l,  we  obtain 


which  is  equal  to  the  ordinate  vw. 

If  the  load  moves  to  any  point  in  panel  FE,  the  moment  about  G  is 


M"=  a-P(x-  b-d)2. 

If  x  -  b  +  2d  and  P  =  1,  we  obtain 


which  is  equal  to  ordinate  MO,  so  the  influence  line  vu  for  loads  in  panel 
EF  can  be  drawn. 

If  the  load  P  moves  to  any  point  to  the  left  of  E,  the  moment  about 
G  is  M'"=(P*/L)a-P(a-b).  If  x  =  L,  we  obtain  Af'"  =  (PL/L)a  - 
P(L-b)=Pa-Pa  =  Q.  If  x  =  b  +  2d  and  P=l,  we  obtain  M"'  =  (b  + 
2d)a/L  —  2d  =  b/L(a-2d),  which  is  the  same  as  found  above  for  ordi- 
nate uo.  So  the  line  Ou  can  be  drawn,  which  is  the  influence  line  for 
loads  to  the  left  of  E.  Then  we  have  the  complete  influence  line  OuvtO' 
for  moments  about  G.  As  is  seen,  this  influence  line  has  three  different 
slopes  and  consequently  there  will  be  three  different  moment  increments 
to  consider.  Let  W\  represent  the  resultant  of  the  loads  to  the  left  of 
E,  W2  the  resultant  of  the  loads  in  panel  EF,  W3  the  resultant  of  the 
loads  to  the  right  of  F,  and  let  W  represent  the  total  load  on  the  span  ; 
that  is,  W=W1  +  W2+  W3.  Then  for  the  moment  about  G,  we  have 

M  =  yWl  +  ylW2  +  y2W3    ...........................  .  .  (1). 

Now  if  all  loads  move  to  the  right  a  distance  dx,  we  have 

AM=  Wldxtanpl  +  W2tan/32  -  W3tan(33    ................  (2), 

for  the  increment  of  the  bending  moment  about  G.  If  the  loads  are  in  the 
position  for  maximum  moment  about  G,  this  increment  will  be  equal  to 
zero  and,  hence,  (2)  would  become 


DESIGN  OF  SIMPLE  RAILED  AD  BRIDGES  499 

0  =  Wltanpl  +  W2tan/32  -  W3tanp3    ....................  (3). 


But, 

b  +  d\  a  +  2b      L  +  b 


Substituting  these  values  of  the  tangents  in  (3),  we  have 


Ll  Li  ft 

Substituting  L-a  for  b,  we  have 


from  which  we  obtain 

W1  +  2W2  _W 
a  ~  L" 

That  is,  the  load  to  the  left  of  E  plus  twice  the  load  in  the  panel  EF 
divided  by  the  distance  a  must  equal  the  total  load  on  the  span  divided 
by  the  length  of  the  span  when  the  maximum  stress  in  top  chord  eg 
occurs. 

W  remaining  constant,  the  increment  of  the  bending  moment  about 
G  can  change  signs  only  as  a  load  passes  joint  F,  hence  there  will  be  a 
load  at  F  when  the  maximum  live-load  stress  in  eg  occurs.  Then  by 
placing  the  live  load  according  to  the  above  equation  (4)  '  with  a  load  at 

F,  and  taking  moments  about  G  (considering  the  part  of  the  structure 
to  the  left  of  section  6-6)   the  maximum  live-load  stress  in  eg  can  be 
readily  determined. 

The  reaction  at  A  would  be  determined  first,  then  the  concentration 
at  E,  due  to  the  loads  in  panel  EF,  which  can  be  done  by  taking  moments 
about  F.  Let  R  —  reaction  at  A,  r  =  the  concentration  at  E,  due  to  the 
loads  in  panel  EF,  and  m  =  the  moment  of  all  loads  to  the  left  of  E  about 

G.  Then  taking  moments  about  G,  we  have 


Dividing  this  moment  by  h,  the  height  of  post  gG,  we  obtain  the  hori- 
zontal component  of  the  stress  in  top  chord  eg  and  multiplying  this  com- 
ponent by  sec<£2,  we  obtain  the  maximum  live-load  stress  in  top  chord  eg. 

The  stress  in  top  chord  ce  is  determined  in  a  similar  manner.  The 
above  equation  (4)'  would  be  applied  to  determine  the  position  of  the 
loading.  The  value  of  a  would  be  taken  in  that  case  equal  to  the  dis- 
tance AE. 

208.  Graphical  Determination  of  Live-Load  Stresses  in  Pettit 
Trusses.  —  The  influence  line  method  outlined  in  Art.  103  is  the  most  con- 
venient graphical  method  to  use.  The  work  as  a  whole  is  practically 
the  same  as  shown  in  Art.  205  for  curved  chord  Pratt  trusses. 

As  an  illustration,  let  it  be  required  to  determine  the  live-load  stress 
in  top  chord  eg  of  the  truss  shown  at  (a)  in  Fig.  342.  The  first  thing 
to  do  is  to  draw  the  influence  line  for  reaction,  as  shown  at  (b).  The 


500 


STRUCTURAL  ENGINEERING 


stress  in  eg  can  be  determined  by  taking  moments  about  G  and  consider- 
ing the  part  of  the  structure  to  the  left  of  section  2-2.  Then  evidently 
the  influence  line  for  the  stress  in  eg  will  be  of  the  form  shown  at  (c). 
By  placing  a  unit  load  at  G  and  drawing  ec  (at  (b))  and  ed  parallel, 
respectively,  to  SS'  and  SG,  we  have  the  stress  in  top  chord  eg,  due  to 
the  unit  load  at  G,  represented  by  the  line  ec.  Then  drawing  AB,  at 


Fig.   342 

(c),  and  laying  off  efcr  —  ec  and  drawing  e'A  and  e'B,  we  obtain  the 
influence  line  A  e'B  for  the  stress  in  top  chord  eg.  Then  placing  the  live 
;oad  on  the  span  so  that  there  is  a  load  at  G  and  the  unit  load  to  the  left 
of  G  equal  to  the  unit  load  on  the  span,  and  multiplying  the  loads  PT 
their  respective  ordinates,  as  previously  explained,  the  maximum  live- 
load  stress  in  eg  is  obtained.  In  case  an  equivalent  uniform  live  load  be 
'ised,  the  maximum  stress  in  eg  will  be  obtained  by  multiplying  the  area 


DESIGN  OF  SIMPLE  EAILEOAD  BRIDGES 


501 


of  triangle  Ae'B  by  the  uniform  load  per  foot,  as  previously  explained. 
The  stress  in  bottom  chord  EG  can  be  determined  by  taking  moments 
about  e.  Then  evidently  the  influence  line  for  the  stress  in  EG  will  be 
partly  of  the  form  Ck'D,  shown  at  (d).  But  we  found  in  Art.  207  (see 
Fig.  340)  that  for  moments  the  line  Ck'n  was  a  straight  line  and  undoubt- 
edly Ck'n  will  be  a  straight  line  if  the  influence  line  be  constructed  for 


U3 


Fig.   343 


the  stress  in  EG.  So  if  k'm'  were  known,  Jc'D  could  be  drawn,  thus 
locating  point  o,  and  next  Cn  could  be  drawn  and  then  the  line  on  could 
be  drawn  and  thus  we  would  have  the  complete  influence  line  CnoD  for 
the  stress  in  bottom  chord  EG.  As  is  seen,  the  only  thing  needed  to  con- 
struct this  influence  line  is  the  ordinate  k'm'.  By  placing  a  unit  load  at 
E  and  taking  moments  about  e  (considering  the  part  of  the  structure  to 
the  left  of  section  2-2)  and  dividing  by  the  height  of  post  eE  the  stress 


502  STRUCTURAL  ENGINEERING 

in  bottom  chord  EG,  due  to  the  unit  load  at  E,  is  obtained,  which  is  the 
desired  value  of  k'm'.  By  drawing  Ae  and  eN  we  have  the  truss  AeNEA. 
Now  it  is  readily  seen  that  a  unit  load  at  E  will  produce  the  same  stress 
in  AN  as  would  be  found  in  EG  by  taking  moments  about  e.  By  draw- 
ing km  at  (b),  parallel  to  Ae,  we  have  the  stress  in  AN  represented  by 
the  line  km.  Then  by  laying  off  k'm'  =  km,  the  influence  line  CnoD  can 
be  drawn  as  explained  above.  Then  placing  the  live  load  on  the  span 
in  accordance  with  (4)  of  the  last  article,  and  multiplying  the  loads  by 
the  proper  ordinates  to  the  influence  line  at  (d),  the  maximum  live-load 
stress  in  bottom  chord  EG  will  be  obtained.  The  other  influence  lines 
shown  in  Fig.  342  are  considered  self-explanatory,  provided  Art.  205  is 
thoroughly  understood. 

209.  Remarks  Concerning  Pettit  Trusses. — Pettit  trusses,  as  pre- 
viously stated,  are  used   for  long  span  bridges.      These  trusses   should 
have  economic  heights,  in  which  case,  if  single  paneling  were  used,  the 
floor  system  would  be  very  heavy,  or  the  diagonals  would  have  an  un- 
economic and  awkward  slope.     These  undesirable  features  are  eliminated 
by  the  sub-paneling. 

All  diagonals  and  sub-diagonals  can  be  made  out-and-out  tension 
members  (eye-bars)  if  the  sub-paneling  be  at  the  top  chord.  This  will 
result  in  a  lighter  truss  than  if  the  paneling  be  at  the  bottom  chord.  But 
experience  seems  to  indicate  that  trusses  with  sub-paneling  at  the  bot- 
tom chord  are  more  rigid  than  trusses  with  sub-paneling  at  the  top  chord. 
It  is  a  question  whether  the  lack  of  rigidity,  however,  is  not  due  more 
to  poor  details  than  to  the  form  of  the  truss. 

The  stress  sheets  and  the  detail  drawings  for  Pettit  trusses  are 
gotten  out  in  the  same  manner  as  previously  explained  for  parallel  and 
curved  chord  bridges  and  the  calculations  of  the  details  are  similar  to 
those  previously  shown  for  those  bridges. 

A  fair  idea  of  the  details  of  a  Pettit  truss  can  be  obtained  from 
Fig.  343,  where  the  details  of  a  panel  of  the  bridge  indicated  in  Fig.  339 
are  shown.  These  details  are  of  the  Bismarck  bridge  designed  by  Ralph 
Modjeski  and  built  by  the  American  Bridge  Company.  It  is  a  Northern 
Pacific  Railway  bridge  over  the  Missouri  River  at  Bismarck,  North 
Dakota. 

The  designing  of  the  floor  and  lateral  systems  and  end  bearings  of 
Pettit  truss  bridges  is  exactly  the  same  as  for  the  curved  chord  Pratt 
truss  bridges,  previously  given. 

MISCELLANEOUS  TRUSSES 

210.  Warren  Trusses. — The  truss  shown  in  Fig.  344  is  known  as 
the  Warren  truss.     In  case  of  a  through  bridge,  there  would  be  a  floor 
beam  at  each  of  the  joints  B,  C,  and  D. 

Dead-Load  Stresses.  Let  P  represent  the  dead  load  per  panel  at 
each  of  the  upper  joints  and  W  the  dead  load  per  panel  at  each  of  the 
lower  joints  and  let  R  represent  the  end  reaction  due  to  these  loads. 
The  dead-load  stresses  in  the  web  members  are  indicated  at  (a)  in  Fig. 
344.  These  expressions  can  be  written  from  mere  inspection,  As  is 
readily  seen, 


DESIGN  OF  SIMPLE  RAILROAD  BEIDGES 


503 


Taking  moments  about  b  of  the  forces  to  the  left  of  that  joint  and 
dividing  this  moment  by  h,  we  obtain 


for  the  stress  in  bottom  chord  AB.     Taking  moments  about  B  of  the 
forces  to  the  left  of  that  joint,  we  obtain 


for  the  stress  in  top  chord  be. 

,P  ,P  IP 

\h  +3(yr+  f>)fon0  I  c  +  ^/,y  PJ  fone  \cf 
/^T~     ~~W%. 


L. 


In  a  similar  manner,  taking  moments  about  c,  we  obtain 


for  the  stress  in  bottom  chord  BC  and  by  taking  moments  about  C,  we 
obtain 


for  the  stress  in  top  chord  cd. 

From  this  it  is  seen  that  the  dead-load  stress  in  any  Warren  truss 
is  easily  determined. 

Live-Load  Stresses.  In  case  the  live  load  be  a  uniform  load,  the 
stresses  are  very  easily  determined.  Suppose  the  live  load  to  be  w 
pounds  per  foot  of  truss.  Then  we  have 

2dxw  =  W 

for  the  live-load  panel  load. 

Loading  joint  D   (alone),  we  have 

W 


for  the  maximum  live-load  compression  in  diagonal  dC  (as  indicated  at 
(b)).     Loading  joints  7)  and  C,  we  have 

,  3-fsecO  ".       ..;,:," 

for  the  maximum  live-load  stress  in  each  of  the  diagonals   Cc  and  cB, 
tension  in  cC  and  compression  in  cB. 


504  STRUCTURAL  ENGINEERING 

Loading  joints  D,  C,  and  B,  we  have 


for  the  maximum  live-load  stress  in  diagonal  bB  and  end  post  bA,  tension 
in  bB  and  compression  in  bA. 

The  chord  stresses  will  be  a  maximum  when  the  span  is  fully  loaded  ; 
that  is,  when  joints  D,  C,  and  B  are  loaded.  The  reaction  at  A  is  then 
equal  to  \\W  .  Then  taking  moments  about  b  we  have 


for  the  maximum  live-load  stress  in  bottom  chord  AB.     Taking  moments 
about  B,  we  have 


for  the  maximum  live-load  stress  in  top  chord  be.     Taking  moments  about 
c,  we  have 


for  the  maximum  live-load  stress  in  bottom  chord  BC.     Taking  moments 
about  joint  C,  we  have 


for  the  maximum  live-load  stress  in  top  chord  cd. 

b  c  d 


Fig.  345 

From  this  it  is  seen  that  the  stresses  in  any  Warren  truss  due  to  a 
uniform  live  load  are  easily  determined. 

The  stresses  in  Warren  trusses,  due  to  wheel  loads,  are  very  readily 
determined.  As  an  example,  let  it  be  required  to  determine  the  maximum 
live-load  tensile  stress  in  diagonal  cC.  The  span  would  be  loaded  as 
shown  in  Fig.  345.  Let  P  represent  the  load  in  panel  BC,  the  center  of 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  5Q5 

gravity  of  which  is  z  distance  from  C;  and  let  W  represent  the  total  load 
on  the  bridge,  the  center  of  gravity  of  which  is  x  distance  from  E.  Then 
by  taking  moments  about  E,  we  obtain 

Wx 

R=^7 

for  the  reaction  at  A  and  taking  moments  about  C,  we  obtain 

_P£ 

~2d 

for  the  concentration  at  B. 

Then  for  the  live-load  tensile  stress  in  cC,  we  have 

tWx       Pz\ 
S=  (R-r)SGcO  =  (-J-  -  ^jsecfl.  .  ....................  (1). 

Now  suppose  the  loads  move  a  very  short  distance  to  the  right  or 
left,  say  to  the  left,  we  have 


Wx 

secO-  +  -         sec* 


for  the  stress  in  cC. 

Subtracting  (1)  from  this  last  equation,  we  obtain 


for  the  increment  of  the  stress  in  cC.  Now  this  increment  would  be  zero 
if  the  stress  in  cC  were  a  maximum  and  hence  the  last  equation  would 
become 


But  A,r  =  A#,  as  is  readily  seen,  so  we  have 
W      P       W       P 


when  the  stress  in  diagonal  cC  is  a  maximum.  Expressing  this  in  words, 
the  unit  load  on  the  bridge  is  equal  to  the  unit  load  in  panel  BC  when 
the  maximum  live-load  tensile  stress  in  diagonal  cC  occurs.  A  load  will 
be  at  C.  The  maximum  compression  in  diagonal  cB,  which  is  equal  to 
the  tension  in  cC,  occurs  at  the  same  time.  To  obtain  maximum  tension 
in  bB  and  maximum  compression  in  bA,  the  span  would  be  loaded  from 
the  right  with  a  wheel  at  B  and  panel  AB  loaded  according  to  the  above 
equation  (2). 

Let  it  be  required  to  determine  the  maximum  live-load  stress  in  bot- 
tom chord  BC.  The  span  would  be  loaded  as  shown  in  Fig.  346.  Let 
W  represent  the  total  load  on  the  bridge,  the  center  of  gravity  of  which 
is  x  distance  from  E.  Let  PI  represent  the  load  to  the  left  of  B,  the  cen- 
ter of  gravity  of  which  is  z  distance  from  B,  and  let  P2  represent  the 


506 


STRUCTURAL  ENGINEERING 


load  in  panel  BC,  the  center  of  gravity  of  which  is  y  distance  from  C. 
Then  we  have 


Taking  moments  about  c,  we  obtain 


for  the  stress  in  Z?T. 

c/         c/ 


have 


Fig.   346 

Now,  suppose  the  loads  all  move  a  short  distance  A.r,  then  we  would 


fcr  the  increment  of  the  stress  in  chord  BC.     But  if  the  stress  in  BC 
were  a  maximum,  this  increment  would  be  zero  and  we  would  have 


But  A,r  =  A2  =  A2/,  so  this  last  equation  reduces  to 


from  which  we  obtain 

W 
L 


(3) 


This  last  equation  shows  clearly  how  the  loads  should  be  placed 
for  maximum  stress  in  bottom  chord  BC.  The  case  of  the  other  bottom 
chords  is  similar.  The  stress  in  the  top  chords  is  obtained  by  taking 
moments  about  the  bottom  chord  joints.  The  placing  of  the  loads  in  that 
case  is  simply  a  matter  of  applying  (5)  of  Art.  91. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


507 


The  influence  line  for  stress  in  diagonal  cC  is  shown  in  Fig.  345  and 
the  influence  line  for  stress  in  bottom  chord  BC  is  shown  in  Fig.  340. 
These  are  readily  understood;  in  fact,  the  drawing  of  the  influence  line 


Fig.  347 

for  any  of  the  members  in  a  Warren  truss  is  a  simple  problem,,  provided 
the  work  previously  given  on  influence  lines  is  understood. 

There  are  very  few  out-and-out  Warren  trusses  (such  as  shown  in 
Fig.  344)  built.  They  usually  have  vertical  posts  and  hangers,  as  shown 
in  Fig.  347.  In  that  case  the  loading  for  maximum  stress  and  the  determi- 
nation of  the  stresses  are  the  same  as  for  an  ordinary  Pratt  truss. 

Referring  to  Fig.  347,  the  stress  in  hangers  dD  and  fF  is  the  same  as 
in  hangers  bB  and  hH. 

The  posts  cC,  eE  and  gG  in  the  case  of  through  bridges  have  only 
dead-load  stress,  which  is  due  to  the  load  at  the  top  chord  joints.  To  obtain 
the  maximum  live-load  stress  in  bottom  chord  CE,  due  to  wheel  loads,  a 
wheel  would  be  placed  at  D,  such  that  the  average  unit  load  to  the  left  of 
D  would  equal  the  average  unit  load  on  the  bridge.  The  stress  would  then 
be  obtained  by  taking  moments  about  d.  For  maximum  stress  in  top  chord 
df  the  span  would  be  loaded  in  reference  to  joint  E. 

That  is,  a  wheel  would  be  placed  at  E,  such  that  the  average  unit 
load  to  the  left  of  E  would  be  equal  to  the  average  unit  load  on  the 
bridge.  The  stress  in  df  would  then  be  determined  by  taking  moments 
about  E. 

212.  Double-System  Warren  Truss. — The  truss  shown  at  (a), 
Fig.  348,  is  known  as  a  "double-system"  Warren  truss.  The  dead-load 
stresses  and  stresses  due  to  a  uniform  live  load  can  be  determined  very 
readily  by  considering  the  truss  composed  of  two  independent  trusses, 
one  of  which  is  shown  at  (b)  and  the  other  one  at  (c). 

The  stresses  are  then  determined  in  each  of  these  trusses  and  com- 
bined ;  thus  the  stress  in  the  structure  as  a  whole  is  obtained. 

Dead-Load  Stresses.  Let  P  represent  the  dead  load  per  panel  at 
each  top  joint  and  let  W  represent  the  dead  load  per  panel  at  each 
bottom  joint.  The  end  reaction  on  the  truss  shown  at  (b)  is 


Then   the   dead-load    stress    in   the    web    members    is    as    indicated. 
Taking  moments  about  b,  we  have 

d 
h 

for  the  stress  in  bottom  chord  AC.     Taking  moments  about  C,  we  have 


508  STRUCTURAL  ENGINEERING 

for  the  stress  in  top  chord  bd.     Taking  moments  about  d,  we  have 

S2  = 


The  dead-load  stresses  indicated  on  the 


for  the  stress  in  bottom  chord  CE.     Taking  moments  about  E,  we  have 


for  the  stress  in  top  chord  de. 
truss  shown  at  (c)  are 
obtained  in  a  similar 
manner.  By  combining 
the  stresses  given  on 
the  top  and  bottom 
chords  and  end  posts 
at  (b)  and  (c)  the 
dead  -  load  stresses  in 
the  truss  as  a  whole 
are  obtained. 

Live-Load  Stresses. 
• — Let  w  represent  a 
uniform  live  load  per 
foot  of  truss.  Then 
we  have 

wd=W 

for  the  live  load  per 
panel.  Referring  to 
the  truss  shown  at  (d) 
and  considering  ^W  at 
//  and  W  at  G,  we  ob- 
tain 


sec0 


for  the  maximum  live- 
load  compressive  stress 
in  diagonal  Ef,  and  by 
considering  \W  at  H 
and  W  at  each  of  the 
joints  G  and  E,  we  ob- 
tain 6£(JF/8)sec0  for  the  maximum  live  stress  in  each  of  the  diagonals  dE 
and  dC,  tension  in  dE,  and  compression  in  dC.  Considering  \W  at  H 
and  W  at  each  of  the  joints  G,  E,  and  C,  we  obtain  12£(JF/8)sec0  for 
the  maximum  live-load  tensile  stress  in  diagonal  bC.  Considering  \W 
at  H  and  W  at  each  of  the  joints  G,  E,  and  C  and  \W  at  B,  we  obtain 
2W»ec6  for  the  live-load  stress  in  end  post  bA.  When  the  truss  at  (d)  is 
fully  loaded,  that  is,  W  at  each  of  the  joints  C,  E,  and  G,  and  \W  at 
each  of  the  joints  H  and  B,  we  obtain,  by  taking  moments  about  6, 


DESIGN  OF  SIMPLE  EAILKOAD  BEIDGES  5Q9 


for  the  stress  in  bottom  chords  ABC.    Taking  moments  about  C,  we  obtain 


for  the  stress  in  top  chord  bd.     Taking  moments  about  d,  we  obtain 

(2W)3  ~  (JJF)2  |  -  (W)  Y  =4JFtan0 
h  n  n 

for  the  live-load  stress  in  bottom  chord  CE  and  taking  moments  about  E, 
we  obtain  4^WtanO  for  the  live-load  stress  in  top  chord  df.  The  stresses 
shown  at  (e)  are  obtained  in  a  similar  manner.  Then  combining  the 
stresses  given  on  the  top  and  bottom  chords  and  end  post  at  (d)  and  (e) 
the  live-load  stresses  in  the  truss  as  a  whole  are  obtained. 

In  case  wheel  load  be  used,  the  stresses  can  be  determined  most 
readily  by  the  use  of  influence  lines.  As  an  illustration,,  let  it  be  required 
to  determine  the  stress  in  chord  de  (Fig.  349)  due  to  wheel  loads.  First 
construct  the  influence  lines  for  shear  as  shown  at  (b).  E  is  the  center 
of  moments  when  the  system  drawn  in  full  is  considered.  Then  draw- 
ing ms  parallel  to  OE,  the  distance  ns  is  obtained,  which  gives  the  stress 
in  de,  due  to  a  unit  load  at  E.  Then,  laying  off  at  (c)  the  ordinate  b  =  ns, 
the  influence  line  C-4-J5  is  obtained.  D  is  the  center  of  moments  when  the 
dotted  system  is  considered.  Then  drawing  ut  parallel  to  OD  the  dis- 
tance vt  is  obtained,  which  gives  the  stress  in  de  due  to  a  unit  load  at  D. 
Then  laying  off  the  ordinate  a  —  vt,  the  influence  line  C-3-B  is  obtained. 
Let  us  consider  a  single  load  passing  over  the  span  starting  from  K. 
When  it  reaches  joint  H  half  of  it  is  transmitted  to  the  two  systems. 
Then,  evidently,  the  point  half  way  between  the  two  influence  lines  at 
that  point  will  be  on  the  influence  line  for  stress  in  de.  When  the  load 
reaches  joint  G,  the  load  will  be  supported  by  the  system  shown  in  full 
and,  hence,  the  point  6  will  be  on  the  influence  line  for  stress  in  de.  When 
the  load  reaches  joint  F,  the  load  will  be  supported  by  the  dotted  system 
and,  hence,  the  point  5  will  be  on  the  influence  line  for  stress  in  de. 
Tracing  out  in  this  manner,  we  obtain  the  influence  line  B-7-6-5-4-3-2-1-C 
for  stress  in  de.  Then  placing  the  wheel  loads  (mostly  by  trial)  for 
maximum  stress  in  de  and  multiplying  each  load  by  its  respective  ordi- 
nate to  this  zigzag  influence  line  the  maximum  stress  in  de,  due  to  wheel 
loads  is  obtained.  The  maximum  stress  in  the  other  top  chord  members 
is  obtained  in  the  same  manner. 

The  influence  line  for  the  stress  in  bottom  chord  DE  is  shown  at  (d). 
Considering  the  system  drawn  in  full,  the  center  of  moments  in  deter- 
mining the  stress  in  DE  is  at  d.  Drawing  mx  parallel  to  Ad,  the  dis- 
tance nx  is  obtained,  which  gives  the  stress  in  DE,  due  to  a  unit  load  at 
E  and  drawing  or  parallel  to  SA,  the  distance  ko  is  obtained,  which  gives 
the  stress  in  DE,  due  to  a  unit  load  at  C.  Then  laying  off  at  (d)  the 
ordinates  d  and  c  equal,  respectively,  to  nx  and  ko,  the  influence  line 
.E-9-11-F  is  obtained.  Considering  the  dotted  system,  e  would  be  the 
center  of  moments  in  determining  the  stress  in  DE.  Drawing  wy  paral- 


510 


STKUCTUEAL  ENGINEERING 


lei  to  Ae,  the  distance  ey  is  obtained,  which  gives  the  stress  in  DE 
due  to  a  unit  load  at  F,  and  drawing  zc  parallel  to  Ke,  the  distance  vz  is 
obtained,  which  gives  the  stress  in  DE,  due  to  a  unit  load  at  D.  Now, 
laying  off  at  (d)  the  ordinates  /  and  e  equal,  respectively,  to  ey  and  vz, 
the  influence  line  E-W-12-F  is  obtained.  Then  by  considering  a  single 
load  to  move  over  the  span  as  in  the  above  case,  the  influence  line 
F-14-13-12-11-10-9-8-JB  for  the  stress  in  bottom  chord  DE  is  obtained. 
By  placing  the  wheel  loads  (by  trial)  for  maximum  stress  in  DE  and 


multiplying  each  load  by  its  respective  ordinate  to  this  influence  line,  the 
maximum  live-load  stress  in  bottom  chord  DE  will  be  obtained.  The 
stress  in  any  of  the  other  bottom  chord  members  can  be  determined  in  the 
same  manner. 

Let  it  be  required  to  determine  the  stress  in  diagonal  dE.  Draw- 
ing am  parallel  to  diagonal  dE,  we  have  the  stress  in  the  diagonal  due 
to  a  unit  load  at  E,  and  drawing  bh  parallel  to  dC,  we  have  the  stress 
in  diagonal  dC,  also  diagonal  dE,  due  to  a  unit  load  at  C.  Then  laying 
off  at  (e)  the  ordinates  a'm'  and  b'h'  equal,  respectively,  to  am  and 


DESIGN  OF  SIMPLE  RAILROAD  BEIDGES 


511 


bh,  the  influence  line  H-m'-h'-G  is  obtained.  Then  by  considering  a 
single  load  to  move  over  the  span,  the  influence  line  //-20-19-18-m'-16- 
h'-15-G  for  the  stress  in  diagonal  dE  is  readily  drawn.  Then  by  placing 
the  wheel  loads  (by  trial)  on  the  part  of  this  influence  line  to  the  right 
of  16  and  multiplying  each  load  by  its  respective  ordinate  the  maximum 


Fig.  350 

tensile  stress  in  diagonal  dE  is  obtained.  At  the  same  time  the  maximum 
compression  in  diagonal  dC  is  obtained,  as  the  stress  in  dE  and  dC  are 
equal  and  opposite.  To  obtain  the  maximum  compression  in  dE  (also 
maximum  tension  in  dC)  the  wheel  loads  would  be  placed  to  the  left  of 
16  and  each  load  multiplied  by  its  respective  ordinate.  The  stress  in 
any  of  the  other  web  members  due  to  wheel  loads  can  be  determined  in 
the  same  manner. 


512  STBUCTURAL  ENGINEEEING 

In  order  to  obtain  short  panels,  double-system  Warren  trusses  are 
sometimes  sub-paneled  as  shown  at  (a),  Fig.  350.  The  stresses  are 
determined  in  very  much  the  same  manner  as  shown  above  for  the  trusses 
without  sub-paneling. 

Considering  the  case  of  dead  load,  let  P  represent  the  panel  load  at 
each  upper  panel  point  and  W  the  load  at  each  lower  panel  point.  We 
can  consider  the  sub-paneling  as  being  independent  sub-trusses  or  trussed 
stringers  as  indicated  at  (b)  and  that  the  load  from  these  sub-trusses 
is  transmitted  directly  to  the  main  lower  panel  points,  in  which  case 
there  would  be  2W  on  each. 

Then  by  considering  the  truss  to  be  composed  of  two  separate  trusses 
as  shown  in  Fig.  348  the  stress  in  all  the  main  members  can  be  readily 
determined  and  then  considering  the  additional  stress  in  each  bottom 
chord  and  in  the  lower  half  of  each  diagonal  due  to  the  sub-trusses  the 
dead-load  stresses  throughout  the  truss  are  obtained. 

Let  it  be  required  to  determine  the  dead-load  stress  in  diagonal  eo. 
The  truss  being  symmetrical  about  I  one-half  of  the  load  at  that  point 
can  be  considered  as  transmitted  by  eo  and  also  the  load  P  at  e.  Then 
we  have  (JF  +  P)sec0  for  the  dead-load  stress  in  eo.  This  same  stress 
will  be  in  oE  and,  in  addition,  the  load  W  at  F  will  produce  a  compressive 
stress  of  ^WsecO  in  the  member  which  should  be  added  to  the  above.  So 
we  have  (W  +  P)sec6  +  ^WsecO  for  the  total  dead-load  stress  in  diag- 
onal oE. 

For  the  dead-load  stress  in  diagonal  od,  we  have  ^PsecO  from  panel 
point  /  and  2Wsec6  from  panel  point  G.  So  for  the  total  dead-load 
stress  in  od,  we  have  2Wsec6 +  ^PsecO.  This  same  stress  would  be  in  oG 
(so  to  speak)  but  the  load  W  at  F  would  produce  a  compressive  stress  of 
^WsecO,  which  should  be  subtracted  from  the  above  and  hence  we  obtain 
2WsecQ  +  ^PsecO-^WsecO  for  the  total  dead-load  stress  in  diagonal  oG. 

The  dead-load  stress  in  each  of  the  hangers  bB,  mD,  oF,  etc.,  is  equal 
to  W.  The  stress  in  the  bottom  chord  of  each  sub-truss,  as  CmE,  is  equal 
to  ^WtanO.  This  must  be  added  to  the  stress  found  in  each  bottom  chord 
when  considering  the  truss  as  composed  of  two  independent  trusses. 
That  is,  the  stress  in  each  bottom  chord  is  determined  by  considering  the 
truss  to  be  composed  of  two  independent  trusses  (2W  at  each  lower  joint), 
just  the  same  as  shown  above  in  the  case  of  trusses  without  sub-paneling, 
and  then  the  stress  QJFtanfl)  in  the  bottom  chord  of  the  sub-truss  is 
added  to  this. 

The  stress  in  the  top  chords  and  upper  half  of  the  diagonals  is  just 
the  same  as  if  there  were  no  sub-paneling. 

The  stress  in  be  (the  upper  half  of  the  end  post)  is  equal  to  the 
shear  in  panel  AC  multiplied  by  sec0.  So  we  have  (7JF  +  3^P)  sec0  for 
the  dead-load  stress  in  be.  This  same  stress  occurs  in  Ab  and  an  addi- 
tional amount  of  ^WsecO,  due  to  the  load  at  B.  So  for  the  stress  in  Ab, 
we  have  (7W+  3iP)sec0  +  ^JFsec0.  For  the  dead-load  stress  in  bottom 
chord  AC,  we  have  (7JF  +  3£P)tan0  +  i*Ftan0.  The  part,  $WtanO,  is 
due  to  the  load  W  at  B. 

In  determining  the  stress  in  the  hanger  cC  the  load  at  c,  E,  e,  and 
7  can  be  considered  as  not  affecting  that  member.  Then  we  have  \P 
from  /,  2W  from  G,  P  from  d  and  2W  from  C,  making  in  all  (4fF  f  1JP), 
which  can  be  considered  to  be  the  stress  in  cC. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  513 

In  determining  the  live-load  stresses  the  work  is  greatly  simplified 
by  using  an  equivalent  uniform  load. 

Let  w  represent  the  live  load  per  foot  of  span.  Then  we  have  wd  —  W 
for  live  load  per  panel.  As  an  example,  let  it  be  required  to  deter- 
mine the  maximum  live-load  tension  in  diagonal  od  (shown  at  (a),,  Fig. 
350).  We  would  load  all  panels  from  P  to  F  (inclusive).  This  would  be 
equivalent  to  placing  2W  at  each  of  the  joints  0,  M,  K,  I,  and  G  and  \W 
at  E.  The  load  at  0,  K,  and  G  are  the  only  ones  that  produce  stress  in 
od.  So  we  have  [(W)l/8  +  (2W)3/8  +  (2W)5/8]sccO  for  the  maximum 
live-load  tension  in  diagonal  od. 

In  determining  the  maximum  live-load  tensile  stress  in  diagonal  oG 
the  load  at  F  would  be  omitted  as  it  would  cause  compression  in  that 
member.  That  is,  the  panels  from  P  to  G  (inclusive)  would  be  loaded 
which  would  be  equivalent  to  placing  2W  at  each  of  the  joints  0,  M,  K, 
and  7  and  \^W  at  G.  Then  ignoring  the  load  at  I  and  M,  we  have 
[(W)l/8+  (2W)3/8+  (l|JF)5/8]sec0  for  the  maximum  live-load  tensile 
stress  in  diagonal  oG. 

To  determine  the  maximum  live-load  compression  in  od,  panel  points 
B,  C,  and  D  would  be  loaded,  in  which  case  the  stress  in  od  would 
be  l/8(2JF)sec0,  which  is  due  to  the  2W  at  C. 

In  the  case  of  diagonal  oG,  the  maximum  compression  in  it  would 
occur  when  panels  B  to  F  (inclusive)  were  loaded. 

The  load  at  E  would  produce  no  stress  in  oG.  One-half  of  the  load 
at  F  would  be  transmitted  to  G.  This  would  produce  a  compressive 
stress  of  ^WsecO  in  oG  but  five-eighths  of  this  would  be  transmitted  back, 
so  we  would  have  (J?Fsec#)f  for  the  compressive  stress  in  oG  due  to  the 
load  W  at  F.  The  load  2W  at  C  is  the  only  load,  except  the  one  at  F, 
that  affects  diagonal  oG.  So  we  have  [1/8W  +  (JJF)3/8] sec0  for  the 
maximum  live-load  compressive  stress  in  diagonal  oG. 

The  maximum  live-load  tensile  and  compressive  stresses  in  any  of 
the  diagonals  due  to  a  uniform  live  load  can  be  determined  in  the  same 
manner  as  shown  above  for  diagonals  oG,  od,  eo,  and  oE.  The  maximum 
stress  in  the  chord  members  due  to  a  uniform  live  load  occurs  when  all  the 
joints  from  B  to  P  (inclusive)  are  loaded.  The  sub-paneling  is  first 
ignored  and  a  load  of  2W  is  assumed  to  be  at  each  of  the  joints  C,  E,  G, 
I,  K,  M,  and  0.  Then  the  stresses  in  the  chords  are  determined  by  con- 
sidering the  truss  to  be  composed  of  two  independent  trusses  as  previously 
explained.  But  to  the  stress  in  each  bottom  chord  thus  obtained  the 
stress  ^WtanO  must  be  added.  ^Wtan.6  is  the  stress  in  the  bottom  chord 
of  each  sub-truss  as  CmE,  EoG,  etc. 

In  case  wheel  loads  be  used,  the  stresses  can  be  determined  most 
readily  by  the  use  of  influence  lines.  The  influence  lines  are  easily  drawn, 
some  of  which  are  shown  in  Fig.  350.  The  influence  lines  for  the  top 
chords  are  the  same  as  if  the  truss  were  not  sub-paneled.  The  influence 
line  E-8-9-10-15-11-12-13-14-F,  shown  at  (d),  is  for  bottom  chord  GI. 
The  influence  line  G-lS-^-lG-m-lS-lO-SO-H,  shown  at  (e),  is  for  diagonal 
en,  and  G-15-fc-16-r-ra-18-19-20-#  is  for  diagonal  nl.  The  influence 
line  X-3-4-5-7-8-9-10-M,  shown  at  (f),  is  for  diagonal  nf  and  #-3-4-5-6- 
7-8-9-10-M  is  for  diagonal  Gn.  The  construction  of  these  lines  will  be 
readily  understood  by  referring  to  the  diagrams  at  (c)  and  (g). 


514 


STRUCTURAL  ENGINEERING 


213.  Whipple  Trusses. — The  truss  shown  at  (a),  Fig.  351,  is 
known  as  a  Whipple  truss.  Dead-load  stresses  and  stresses  due  to  uni- 
form live  load  in  this  type  of  truss  are  readily  determined  by  considering 
the  truss  as  composed  of  two  independent  trusses.  In  determining  the 
dead-load  stresses  in  the  truss  shown  at  (a)  the  truss  can  be  considered 
as  composed  of  two  independent  trusses,  one  of  which  is  shown  at  (b) 
and  the  other  one  at  (c).  Let  W  represent  the  dead  load  per  panel,  one- 


third  applied  at  each  top  joint  and  two-thirds  at  each  bottom  joint.     Tak 
ing  moments  about  b  and  considering  the  truss  at  (b),  we  obtain 


for  the  stress  in  CE.     Likewise  taking  moments  about  a  and  considering 
the  truss  at  (c),  we  obtain 

-- 


DESIGN  OF  SIMPLE  EAILROAD  BRIDGES 


51  fi 


for  the  stress  in  BD.     Then  the  stress  in  bottom  chord  CD  is  equal  to 


U^W  \  tan<£  +  2JFtan<£  =  G| 
Again  taking  moments  about  E  and  considering  the  truss  at  (b)  we  obtain 

(- 

V  ^ 


for  the  stress   in  bd  and  taking  moments   about  D   and  considering  the 
truss  at  (c),  we  obtain 


for  the  stress  in  ac.     Then  we  have 


^  W  ) 

&       I 


=  11     Wtan<j> 


for  the  dead-load  stress  in  top  chord  be.     The  dead-load  stress  in  any 
chord  member  can  be  readily  determined  in  this  manner. 


Fig.   352 


Considering  the  trusses  at  (b)  and  (c)  the  dead-load  stresses  in 
the  web  members,  as  indicated,  are  readily  determined. 

Let  P  represent  the  uniform  live  load  per  panel.  The  live-load 
stresses  in  the  chord  members  will  then  be  as  indicated  on  the  trusses 


516 


STRUCTURAL  ENGINEERING 


at  (d)  and  (e).  These  are  determined  in  exactly  the  same  manner  as 
explained  above  for  the  dead-load  stresses  in  the  chords.  The  live-load 
stresses  in  the  web  members  are  indicated  at  (d)  and  (e).  These  are 
determined  as  previously  explained.  For  example,  considering  the  truss 
at  (d)  and  loading  the  panels  from  J  to  E  (inclusive),,  we  obtain 

^ •  (  ~  +  2  +  4  +  G  )  sec<9  =  1  &|jPsec0 

-Lu  y (V  i  ti 

for  the  maximum  live-load  tensile  stress  in  diagonal  bE  and  also 
12  J  x  (P/10)  for  the  maximum  live-load  stress  in  post  bC.  Likewise, 


Fig.  353 

loading  the  panels  from  J  to  F  (inclusive)  and  referring  to  the  truss  at 
(e),  we  obtain 

1 


for  the  maximum  live-load  stress  in  diagonal  cF  and  also  8  J  x  (P/10)  for 
the  maximum  live-load  stress  in  post  cD. 

If  wheel  loads  are  used  the  stresses  can  be  determined  most  readily 
by  use  of  influence  lines.  Influence  lines  for  stresses  in  Whipple  trusses 
are  constructed  practically  in  the  same  manner  as  shown  above  for  the 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


517 


double-system  Warren  trusses.  Some  of  these  influence  lines  are  shown 
in  Fig.  352.  The  one  at  (c)  is  for  top  chord  cd  and  the  one  at  (d)  is 
for  diagonal  cF.  The  construction  and  use  of  influence  lines  in  the 
case  of  the  Whipple  truss  is  quite  simple  provided  the  work  on  influence 
lines  previously  given  is  understood. 


Fig.   354 

214.  Lattice  Truss. — The  truss  shown  at  (a),  Fig.  353,  is  known 
as  a  lattice  truss.  It  can  be  considered  composed  of  four  independent 
trusses,  one  of  which  is  shown  at  (b)  and  another  at  (c)  and  by  turning 
these  two  end  for  end,  we  obtain  the  other  two  trusses.  By  determining 
the  dead  and  maximum  live-load  stress  in  each  member  of  the  trusses 
shown  at  (b)  and  (c)  and  combining  the  stresses,  the  stress  in  the 


518  STRUCTURAL  ENGINEERING 

truss  shown  at  (a)  is  obtained.  The  determination  of  the  dead-load 
stress  is  a  simple  problem  and  the  determination  of  the  live-load  stresses 
is  practically  as  simple  provided  a  uniform  live  load  be  used.  Let  P 
represent  the  uniform  live  load  per  panel.  Considering  the  truss  at  (b) 
and  loading  joint  C  (only)  we  obtain  f  Psec#  for  the  live-load  tension 
in  eC  and  compression  in  eG.  Loading  joint  G  (only)  we  obtain 
f  Psecfl  for  the  live-load  tension  in  eG  and  compression  in  eC.  Loading 
both  joint  C  and  G  we  obtain  -^g0-  Psec<£  for  the  maximum  live-load  ten- 
sion in  bC  and  fPsec#  for  the  maximum  live-load  tension  in  kG,  also 
^  Psee(£  and  f  Psec<£  for  the  live-load  compression  in  b  A  and  kK,  re- 
spectively. Loading  joints  C  and  G  and  taking  moments  about  B,  we 
obtain 


for  the  live-load  stress  in  AC.     Taking  moments  about  C,  we  obtain 


for  the  live-load   stress  in   be.     Taking  moments   about  k    (considering 
reaction  at  K),  we  obtain 


for  the  live-load  stress  in  GK.     Taking  moments  about  e   (considering 
the  forces  to  the  left),  we  obtain 


for  the  live-load  stress  in  CG.  The  stresses  due  to  a  uniform  live  load 
in  the  trusses  shown  at  (b)  and  (c)  are  readily  determined  in  this  man- 
ner. Then  by  combining  these  carefully  for  the  chords  and  end  posts, 
the  stresses  throughout  the  truss  are  obtained. 

In  case  ^yheel  loads  are  used,  the  stress  therefrom  can  be  most 
readily  determined  by  the  use  of  influence  lines.  These  are  constructed 
very  much  the  same  as  previously  shown  for  double-system  Warren  and 
Whipple  trusses. 

Several  railroad  companies  use  the  lattice  truss  shown  at  (a).  The 
details  of  part  of  one  of  these  trusses  are  shown  in  Fig.  354.  These 
details  are  taken  from  the  standard  plans  of  the  Chicago  &  North  West- 
ern Railway  Company  (W.  C.  Armstrong,  engineer  of  bridges). 

DEFLECTION  AND  CAMBER  OF  TRUSSES 

215.  Analytical  Determination  of  Deflection. — Let  it  be  required 
to  determine  the  deflection  of  joint  C  of  the  truss  shown  in  Fig.  355, 
due  to  any  loads.  In  the  case  of  pin-connected  trusses  the  deflection 
is  due  to  the  distortion  of  the  members  and  to  a  small  clearance  between 
the  pins  and  pin  holes,  while  in  the  case  of  riveted  trusses  the  deflection 
is  due  wholly  to  the  distortion  of  the  members.  We  will  first  consider 
the  deflection  of  joint  C,  due  entirely  to  the  distortion  of  the  members. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


519 


Let  u  represent  the  stress  in  any  member  as  aE,  due  to  a  unit  load 
placed  at  C,  and  let  S  represent  the  stress  in  the  same  member  due  to 
any  loads  placed  at  any  panel  points. 


Fig.   355 

Let  L  represent  the  length  of  member  aB  and  A  its  gross  area  of 
cross-section.  Then  for  the  work  done  on  the  member  aB  by  the  stress  u, 
we  have  ux^(uL/AE)  which  must  undoubtedly  be  equal  to  J(Ad)l#, 
where  Ad  represents  the  deflection  of  joint  C,  due  to  the  stress  u.  So 
we  have 


from  which  we  obtain 


Ad  = 


for  the  deflection  of  joint  C,  due  to  stress  u.  Now,  evidently,  the  deflec- 
tion of  joint  C,  due  to  any  stress  S  in  member  aB,  will  be  directly  pro- 
portional to  the  deflection  Ad,  due  to  stress  u.  Then  let  Ay  represent 
the  deflection  of  j  oint  C,  due  to  stress  S  and  we  have 

Ay  _S 
Ad=w' 

and  substituting  (uL/AE)u  for  Ad  and  reducing,  we  obtain 

SuL 
^  =  ~AE 

for  the  deflection  of  joint  C,  due  to  any  stress  S  in  member  aB. 

Now,  evidently,  if  w',  M",  etc.,  represent  the  stress  in  the  other  mem- 
bers of  the  truss,  due  to  a  unit  load  at  C,  and  S',  S" ,  etc.,  the  stress 
in  these  members,  due  to  any  loads  placed  at  any  joints,  and  Ay',  Ay", 
etc.,  the  deflection  of  joint  C,  due  to  the  stresses  £',  S" ,  etc.,  we  can 
write  the  formula 


for  the  total  deflection  of  joint  C,  due  to  any  loads  producing  the  simul- 
taneous stresses  S,  S',  S",  etc. 

By  letting  S  =  stress  in  any  member  due  to  any  loading, 

u  =  stress  in  any  member  due  to  a  unit  load  at  any  point 

considered, 

L  =  length  of  any  member  in  inches, 

A  =  gross  area  of  cross-section  of  any  member  in  square 
inches, 


520 


STKUCTUEAL  ENGINEERING 


the  general  formula  for  deflection  can  be  written  as 
SaZ 


Example  1.  Let  it  be  required  to  determine  the  deflection  of  joint 
L3  of  the  150-ft.  truss  shown  in  Fig.  279,  due  to  the  given  dead  and  live 
loads. 

The  live-load  stresses  given  for  the  truss  members  in  Fig.  279  should 
not  be  used  in  determining  the  deflection  as  they  do  not  occur  simul- 
taneously. The  maximum  deflection  will  occur  when  the  span  is  fully 
loaded  and  a  sufficiently  accurate  result  will  be  obtained  by  using  an 
equivalent  uniform  live  load,  in  which  case  the  live-load  stresses  will  be 
directly  proportional  to  the  dead-load  stresses,  and  hence  they  can 
readily  be  determined  directly  from  the  dead-load  stresses  by  the  use  of 
a  slide  rule. 

The  maximum  live-load  stress  in  the  end  post  (see  Fig.  279)  is 
263,000  Ibs.  Dividing  this  by  the  secant  of  the  slope  of  the  post,  we 
obtain 

263,000  -1.3  -203,000  Ibs. 

for  the  maximum  live-load  shear  in  the  end  panel  LO-L1  (see  Art.  174). 
Then  substituting  this  value  for  S  in  the  formula  of  Art.  123,  we  obtain 


P'=  (203,000) 


-d)  =203,000-62.5  =  3,250  Ibs. 


for  the  equivalent  uniform  live  load  per  foot  of  truss  for  determining 
the  live-load  stress  in  the  web  members,  but  as  the  chords  contribute 
most  to  the  deflection  we  will  use 

3,250  -  3,250  (flft  +  2-5)  %  =  3,120  Ibs. 

per  foot  of  truss  for  the  equivalent  uniform  live  load,  which  is  really 
the  correct  equivalent  uniform  live  load  for  determining  the  maximum 
chord  stresses.  (See  Art.  123.) 


LI 

t>  *    ssooo  # 

i.  +  I&3OOO* 

+^  18000* 


L2  L3 

D+      88000* 

L  f-   26OOOO  * 

+3^8000* 


Fig.   356 


As  seen  from  Fig.  279,  the  dead  load  per  foot  of  truss  is  2,110-7-2  = 
1,055  Ibs.  Then  multiplying  each  of  the  dead-load  stresses  given  for  the 
truss  members  in  Fig.  279  by  f;-J|J  the  live-load  stresses  for  determin- 
ing the  deflection  of  joint  L3  are  obtained.  Then  adding  these  to  the 
dead-load  stresses,  the  total  stress  in  each  truss  member,  as  shown  in 
Fig.  356,  is  obtained. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES  521 

Then  from  Figs.  279  and  356  the  following  table  can  be  computed: 

L      S       A       SL       u     SuL 
Member    Ins.     Lbs.    Sq.  Ins.    EA      Lbs.     EA 

Ul-LO 
LQ-L2 
L2-L3 
U1-U2 
U2-U3 
U1-L2 
U2-L3 
U1-L1 
U2-L2 
U3-L3 

(£  =  30,000,000) 


468 

-337,000 

46.18 

-0.1138 

-0.6500 

0.0739 

600 

+218,000 

23.52 

+0.1853 

+0.4166 

0.0772 

300 

+348,000 

36.98 

+0.0941 

+0.8332 

0.0784 

300 

-348,000 

40.58 

-0.0857 

-0.8332 

0.0714 

300 

-392,000 

45.08 

-0.0869 

-1.2498 

0.1086 

468 

+203,000 

26.48 

+0.1196 

+0.6500 

0.0777 

468 

+67,000 

19.80 

+0.0527 

+0.6500 

0.0343 

360 

+71,000 

14.44 

+0.0590 

0 

0 

360 

-87,000 

19.80 

-0.0527 

-0.5000 

0.0263 

360 

-9,000 

19.80 

-0.0054 

0 

0 

Total  deflection  of  joint  L3  =  1.0996 ins. 

As  L3  is  at  the  center  of  the  span  and  the  bridge  symmetrically 
loaded,  the  deflection  of  L3  can  be  obtained,  as  shown  by  the  above  table, 
by  determining  the  deflection,  due  to  just  half  of  the  truss  members,  and 
then  multiplying  this  deflection  by  two.  In  case  of  the  determination  of 
the  deflection  of  any  other  joint  (except  £73)  all  of  the  truss  members 
would  have  to  be  considered.  For  example,  let  it  be  required  to  deter- 
mine the  deflection  of  joint  L2  (see  Fig.  279).  The  table  in  that  case 
would  be  just  the  same  as  the  above,  except  that  the  last  two  columns 
on  the  right  would  have  two  values  in  each  case  as  the  u  for  the  members 
on  one  side  of  the  center  of  span  would  be  different  from  the  u  for  the 
corresponding  members  on  the  other  side  of  the  center  of  the  span.  For 
example,  placing  a  unit  load  at  L2,  we  have  four-sixths  of  a  pound  for 
the  reaction  at  one  end  of  the  truss,  and  two-sixths  for  the  reaction  at 
the  other  end.  Then  for  the  u  in  one  end  post  we  would  have  $  x  1.3  = 
0.8333,  and  |  x  1.3  =  0.4166  for  the  u  in  the  other  end  post;  and  likewise 
|  x  0.8333  =  0.5555  for  the  u  in  one  bottom  chord  LO-L2,  and  f  x  0.8333  = 
0.2777  for  the  u  in  the  other  bottom  chord  LO-L2.  Thus  it  is  throughout 
the  span.  The  last  two  right-hand  columns  in  the  table  would  simply 
have  double  numbers  and  the  summation  of  the  SuL/AE  would  not.be 
multiplied  by  two.  Otherwise,  the  table  for  each  of  the  other  joints 
would  be  just  the  same  as  shown  above  for  joint  L3. 

In  case  the  loading  causing  the  stresses  S  were  placed  unsym- 
metrically,  each  member  would  be  listed  separately  in  the  tables.  In  that 
case  it  is  advisable  not  to  have  two  or  more  members  with  the  same  mark, 
as  in  the  case  of  the  truss  shown  in  Fig.  279. 

It  is  customary  to  consider  the  distortion  of  compression  members  as 
negative,  and  that  of  the  tension  members  as  positive.  In  that  case  the 
tensile  stresses,  both  u  and  S,  should  be  indicated  as  plus,  and  the  com- 
pressive  stress  as  negative,  as  shown  in  the  above  table. 

In  case  the  longitudinal  displacement  of  any  joint  be  desired,  it 
can  be  determined  in  the  same  manner  as  shown  above  for  deflection, 
except  that  the  unit  load  applied  at  the  joint  in  question  would  be  con- 


522 


STRUCTURAL  ENGINEERING 


sidered  as  acting  horizontally  instead  of  vertically.  In  that  case  it  is 
necessary  to  pay  strict  attention  to  the  signs  of  the  u's  and  S's,  as  they 
may  not  have  like  signs,  which  they  ordinarily  have  in  the  case  of 
deflection. 

As  an  example,  let  it  be  required  to  determine  the  horizontal  dis- 
placement of  joint  L3  of  the  truss  shown  in  Fig.  279,  due  to  the  stresses. 
The  unit  load  would  be  applied  at  L3,  as  shown  in  Fig.  357.  The  end 
Z/6  being  fixed  and  the  end  LO  being  on  rollers,  the  only  members  that 
would  really  be  stressed  by  the  unit  load  would  be  L3-L4  and  L4-L6, 
and  hence  we  need  consider  only  these  members  in  determining  the  hori- 
zontal displacement  of  L3.  The  stress  in  each  of  these  members,  due  to 
the  unit  force,  as  is  obvious,  is  1  Ib.  Then  making  the  following  table, 
the  horizontal  displacement  of  L3  is  obtained: 

L  S  A  SL  u  SuL 

Member         Ins.  Lbs.        Sq.  Ins.          EA  Lbs.  EA 

L3-L4  ...   300       -348,000       36.98       -0.0941       -1.0000       0.0941 
L4-L6  ...   600       -218,000       23.52       -0.1853       -1.0000       0.1853 

Total  horizontal  displacement  of  joint  Z/3  =  0.2794  ins. 

This  shows  that  joint  L3  will  move  about  \"  to  the  left.  It  will  be 
seen  that  this  table  is  really  compiled  from  the  table  given  above  in 
determining  the  deflection  of  L3. 

The  horizontal  displacement  of  any  of  the  other  lower  chord  joints 
can  be  determined  in  a  similar  manner,  and  just  as  readily  after  the 
tables  for  the  deflection  of  those  joints  are  made. 

The  determination  of  the  horizontal  displacement  of  the  top  chord 
joints,  while  similar  to  that  shown  for  the  bottom  chord  joints,  is  not  so 
simple,  owing  to  the  fact  that  more  members  are  involved.  As  an  illus- 


/'  U4 


Fig.   357 

tration,  let  it  be  required  to  determine  the  horizontal  displacement  of 
joint  £74  (Fig.  357).  The  unit  load  would  be  applied  at  £74  as  indicated. 
The  horizontal  reaction  of  this  load  would  be  at  L6,  as  indicated  (the 
end  L6  being  fixed),  but  in  addition,  the  unit  load  at  £74  would  produce 
a  positive  reaction  of  r  at  LO  and  an  equal  negative  reaction  r  at  L6. 
These  vertical  reactions  and  the  1  Ib.  horizontal  reaction  at  L6  would 
have  to  be  considered  in  determining  the  u's  for  the  members  throughout 
the  truss.  For  each  of  the  vertical  reactions  we  have  r=h/L.  After  the 
u's  are  determined  for  the  members  throughout  the  truss,  the  horizontal 
displacement  of  the  top  chord  joints  is  determined  by  making  a  table  for 
each  as  explained  above. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


523 


tf 


(a) 


--'C' 


In  determining  the  deflection  of  pin-connected  bridges,  the  distor- 
tion (SL/AE)  of  each  member  should  be  increased  by  -^  in.  to  allow 
for  the  clearance  between  the  pins  and  pin  holes.  Otherwise  the  work 
of  determining  the  deflection  of  pin-connected  bridges  is  exactly  the  same 
as  shown  above  for  the  riveted  bridge. 

216.  Graphical  Determination  of  Deflection. — Let  ACS  at  (a), 
Fig.  358,  represent  a  cantilever  frame  and  suppose  a  vertical  load  P  be 
applied  at  C.  This  load  will  cause  tension  in  member  AC  and  compres- 
sion in  member  BC  and  hence  the  distortion  in  AC  will  increase  its 
length  while  the  distortion  in  BC  will  decrease  the  length  of  that  member. 
Let  Ca  represent  the  distortion  of  AC  and  Ce  the  distortion  of  BC  and 
let  us  assume  that  the  joints  A,  C,  and  B  are  pin  connected  so  that  the 
members  AC  and  BC  are  free  to  turn  about  their  ends.  Now,  undoubt- 
edly, the  distortion  of  the  members  AC  and  BC  will  cause  the  frame  to 
deflect  to  some  position  as 
AC'B.  As  the  joints  A  ^  I? 

and   B   are   fixed   in   posi-       ^^  ^a 

tion  the  deflection  of  the 
frame  really  consists  of 
joint  C  moving  to  C"  and, 
hence,  the  determination 
of  the  deflection  is  simply 
a  matter  of  locating  C'  in 
reference  to  joints  A  and 
B,  which  remain  fixed.  By 
taking  A  as  a  center  and 
A  a  as  a  radius,  the  arc  «C" 
could  be  described,  and 
taking  B  as  a  center  and 
Be  as  a  radius  the  arc 
eC'  could  be  described 
and  thus  the  point  C' 
would  be  located. 

As  another  example,  let  D-F-G-E  at  (d)  represent  a  cantilever  frame 
supporting  a  vertical  load  P  at  G.  Let  Fd,  Ff,  G'k,  and  Gm  represent 
the  distortions  of  the  members  DF,  FE,  FG,  and  GEf  respectively.  Now, 
owing  to  these  distortions,  the  frame  will  deflect  to  some  position  as 
D-F'-G"-E.  As  is  obvious,  the  problem  of  determining  the  deflection  of 
the  frame  consists  of  locating  F'  and  G"  in  reference  to  the  fixed  joints 
D  and  E.  First,  taking  D  as  a  center  and  Dd  as  a  radius,  the  arc  dF' 
could  be  described ;  and  taking  E  as  a  center  and  Ef  as  a  radius  the  arc 
fF'  could  be  described;  and  thus  the  position  of  F'  would  be  determined, 
which  is  the  deflected  position  of  joint  F.  Now,  having  joint  F  located, 
the  deflected  position  of  joint  G  can  be  determined  in  reference  to  E  and 
F.  Suppose,  for  the  time  being,  that  joint  G  be  disconnected  and  that 
member  FG  moves  to  the  parallel  position  F'G',  while  EG  remains  in 
position:  Then,  taking  F'  as  a  center  and  F'k  as  a  radius,  the  arc  kG" 
could  be  described ;  and  taking  E  as  a  center  and  Em  as  a  radius,  the  arc 
mG"  could  be  described;  and  thus  the  position  of  G"  would  be  deter- 
mined, which  is  the  deflected  position  of  joint  G;  and  hence  the  deflected 
position  of  the  entire  frame  would  be  known. 


•''(e) 


Fig.  358 


524  STRUCTURAL  ENGINEERING 

Now  it  is  evident  that,  theoretically,,  the  deflection  of  any  cantilever 
frame  could  be  determined  in  the  same  manner  as  shown  for  the  above 
two  cases,  and  as  any  truss  can  be  considered  as  being  composed  of  one 
or  more  cantilevers,  the  same  is  true  of  any  truss.  But,  in  practice,  the 
application  of  the  method  is  practically  impossible,  as  the  distortions 
of  the  members  are  so  small  compared  with  their  lengths  that  if  the 
members  be  laid  off  to  a  reasonable  scale  it  would  be  practically  impos- 
sible to  measure  the  distortions  and  deflections.  This  very  fact,  how- 
ever, makes  possible  the  graphical  method  in  general  use,  in  which  it  is 
assumed  that  the  distortions  are  so  small  in  comparison  with  the  lengths 
of  the  members  that  all  such  arcs  as  aC' ,  eC' ',  dF' ,  mG" ,  etc.,  without 
appreciable  error  can  be  considered  straight  lines;  that  is,  the  arc  and 
its  tangent,  in  each  case,  are  assumed  to  coincide.  One  can  satisfy  him- 
self as  to  the  accuracy  of  this  assumption  by  trial.  As  a  suggestion,  sup- 
pose the  distance  AB  (Fig.  358 (a))  be  laid  off  equal  to  10  ft.;  AC  and 
BC  equal  to  15  ft.  and  18.5  ft.,  respectively;  and  Ca  and  Ce  each  equal 
to  -J  of  an  in.  and  the  arcs  AC'  and  eC'  described,  as  explained  above, 
and  the  difference  between  these  arcs  and  their  tangents  noted:  By 
assuming  that  all  such  arcs  as  AC'  and  eC' ,  etc.,  coincide  with  their 
tangents,  the  deflection  of  any  truss  can  be  determined  by  simply  laying 
off  the  distortions  and  drawing  perpendiculars  to  these.  As  an  example, 
let  us  consider  the  case  shown  at  (a),  Fig.  358.  The  joints  A  and  B 
are  considered  fixed.  Let  us  take  any  point  A'  at  (b),  as  origin,  to 
determine  the  deflection  of  the  joints  A,  C,  and  B.  Now,  as  both  A 
and  B  are  fixed,  they  will,  as  regards  deflection,  be  at  A'  and  hence  we 
have  A  and  B  located.  Joint  C  in  reference  to  A  moves  to  the  right  the 
distance  Ca.  Then  at  (b)  lay  off  Ca  from  A'  (the  location  of  joint  A) 
to  the  right  as  shown.  Joint  C  in  reference  to  B  moves  toward  B  the 
distance  Ce.  Then  at  (b)  lay  off  Ce  from  Af  (the  location  of  joint  Z?), 
as  shown.  Then,  by  drawing  from  e  and  a  the  perpendiculars  aCf  and 
eC'  we  have  C'  located,  which  is  the  deflected  position  of  joint  C.  That 
is,  as  Af  is  the  origin  (point  of  zero  deflection),  V  represents  the  deflec- 
tion of  joint  C,  and  h  represents  the  horizontal  movement  of  that  joint. 

Next  let  us  consider  the  case  shown  at  (d),  Fig.  358.  Here  joints 
D  and  E  are  considered  fixed.  Let  Df  at  (e)  be  the  origin  to  determine 
the  deflection  of  joints  D,  E,  F,  and  G.  As  D  and  E  are  fixed,  they  will 
both  be  at  D',  the  origin.  In  reference  to  joint  D  joint  F  moves  to  the 
right  the  distance  Fd.  So  from  D'  (the  location  of  joint  D)  lay  off  Fd 
to  the  right  as  shown.  In  reference  to  E  joint  F  moves  toward  E  the 
distance  Ff.  So  from  D'  (the  location  of  -E)  lay  off  Ff  as  shown.  Then 
by  drawing  the  perpendiculars  dF'  and  fFf  the  point  F'  is  located,  which 
is  the  deflected  position  of  joint  F.  Now  having  F  located,  joint  G  can 
be  located  in  reference  to  joints  F  and  E.  In  reference  to  joint  F  joint  G 
moves  away  from  F  the  distance  G'k,  as  shown.  In  reference  to  joint  E, 
joint  G  moves  toward  E,  or  to  the  left,  the  distance  mG.  So  from  D' 
(the  location  of  joint  E)  lay  off  mG  to  the  left  as  shown.  Then  by  draw- 
ing the  perpendiculars  kG"  and  mG" ',  the  point  G"  is  located,  which  is 
the  deflected  position  of  joint  G.  We  then  have  the  deflection  of  all 
the  joints  determined.  The  distances  V  and  V"  represent  the  deflection 
of  joints  F  and  G,  respectively,  and  h'  and  h"  represent  respectively  the 
horizontal  movement  of  these  joints. 


DESIGN  OF  SIMPLE  RAILROAD  BRIDGES 


525 


As  another  example,  let  it  be  required  to  determine  the  deflection 
of  the  joints  of  the  cantilever  frame  shown  at  (a),  Fig.  359,  where 

Al,  A2 AID  represent  the  distortions  of  the  members.     The  plus 

sign  signifies  that  the  length  of  the  member  is  increased  by  the  distortion, 
and  the  minus  sign  signifies  just  the  opposite.  First,  select  any  point  A' 
at  (b)  as  the  origin.  The  joints  A  and  B  are  fixed  so  they  will  both  be  at 
A' '.  In  reference  to  A,  joint  C  moves  to  the  right  the  distance  Al.  So 
from  A'  (at  (b))  draw  Al  to  the  right.  In  reference  to  B,  joint  C  moves 
toward  B  the  distance  A4.  So  from  B'  (at  (b))  draw  A4  as  shown,  and 
then  drawing  perpendiculars  to  these  distortions  the  point  C'  is  located, 
which  shows  the  deflected  position  of  joint  C.  That  is,  joint  C  has  moved 
from  A'  to  C' .  Now,  having  C  located,  we  can  proceed  to  locate  joint  D 
in  reference  to  joints  B  and  C.  In  reference  to  C,  joint  D  moves  down 
the  distance  A5.  So  from  C'  lay  off  A5  downward  as  shown.  In  refer- 
ence to  B,  joint  D  will  move  toward  B  the  distance  A9.  So  from  B' 
(at  (b))  lay  off  A9  to  the  left.  Then,  drawing  perpendiculars  to  these 
distortions  (A5  and  A9),  we  obtain  D'  which  shows  the  deflected  position 
of  joint  D.  That  is, 
joint  D  has  moved  from 
A'  to  D'.  Now,  having 
both  C  and  D  located, 
we  can  next  locate  the 
joint  E.  In  reference  to 
C,  joint  E  moves  to 
the  right  the  distance 
A2.  So  from  C"  draw 
A2  to  the  right.  In 
reference  to  D,  joint  E 
moves  toward  D  the 
distance  A6.  So  from 
D',  draw  A6  as  shown. 
Then  by  drawing  per- 
pendiculars to  these 
distortions  (A2  and  A6),  we  obtain  E' ,  which  shows  the  deflected  posi- 
tion of  E.  That  is,  joint  E  has  moved  from  Af  to  E'.  Now  having 
joints  E  and  D  located,  we  can  next  locate  joint  F.  In  reference  to  E, 
joint  F  moves  downward  the  distance  A7.  So  from  E'  draw  A  7  down- 
ward. In  reference  to  joint  D,  joint  F  moves  toward  D  the  distance  A10. 
So  from  D'  draw  A10  to  the  left.  Then  by  drawing  perpendiculars  to 
these  distortions  (A7  and  A10),  we  obtain  point  F',  which  shows  the 
deflected  position  of  joint  F.  That  is,  joint  F  has  moved  from  A'  to  F'. 
Now,  having  joints  E  and  F  located,  we  can  locate  joint  G.  In  reference 
to  E,  joint  G  moves  to  the  right  the  distance  A3.  So  from  E'  draw  A3 
to  the  right.  In  reference  to  F,  joint  G  moves  toward  F  the  distance  A8. 
So  from  F'  draw  A8  as  shown.  Then  by  drawing  perpendiculars  to  these 
distortions  (A8  and  A3),  we  obtain  the  point  G' ',  which  shows  the  de- 
flected position  of  joint  G.  That  is,  joint  G  has  moved  from  A'  to  G'. 

Referring  to  the  diagram  at  (b),  we  can  consider  all  of  the  joints 
as  being  at  A'  before  deflection,  and  that  they  deflect  to  the  posi- 
tions indicated.  That  is,  C  moves  from  A'  to  C' ',  D  from  A'  to  D' , 
E  from  A'  to  F',  F  from  A'  to  F',  and  G  from  A'  to  G'.  By  measuring 


(a) 


Fig.  359 


526 


STRUCTURAL  ENGINEERING 


the  vertical  distance  down  from  A'  in  each  case,  the  deflections  are  ob- 
tained and  the  horizontal  movements  are  obtained  by  measuring  the  hori- 
zontal distance  in  each  case  from  a  vertical  line  through  A' . 

As  all  trusses  can  be  considered  as  being  composed  of  one  or  more 
cantilever  frames,  it  is  obvious  that  the  deflection  of  any  truss  can  be 
determined  by  constructing  diagrams  similar  to  those  shown  at  (b)  and 
(e),  Fig.  358,  and  at  (b),  Fig.  359. 

These  diagrams  are  known  as  "Williot  diagrams,"  being  named  after 
the  French  engineer  Williot,  who  proposed  them. 

Example  1.  Let  it  be  required  to  determine  the  deflection  of  the 
truss  shown  in  Fig.  279,  due  to  the  stresses  given  in  Fig.  356.  The 

distortions  of  the  mem- 
bers, due  to  these  stresses 
are  given  in  Example  1  of 
Art.  316  (see  table). 

First,  draw  the  out- 
line of  the  truss  as  shown 
at  (a),  Fig.  360.  The  dis- 
tortions shown  there  for 
one-half  of  the  members 
(they  are  the  same  for  the 
other  half  of  the  truss) 
are  expressed  in  fractional 
form  for  convenience. 

As  the  truss  is  sym- 
metrical about  U3-L3  and 
symmetrically  loaded  the 
deflection  will  be  fully 
known  if  it  be  determined 
for  one-half  of  the  truss. 
By  considering  joint  L3 
as  fixed  in  position  and 
member  U3-L3  as  fixed  in 
direction  either  half  of  the 
truss  can  be  considered  as 
a  cantilever  and  the  deflection  determined  as  explained  above.  Let  us  take 
L3  at  (c)  as  a  starting  point.  Then  the  position  of  joint  U3  is  obtained 
by  laying  off  the  distortion  of  U3-L3  (which  is  j^")  downward,  using  a 
scale  of  iV^^V'.  Then  having  joint  U3  located,  the  position  of  U2 
can  be  determined  in  reference  to  U3  and  L3.  Joint  U2  moves  towards 
U3  and  away  from  L3.  Then  laying  off  the  distortion  of  member 
U2-U3  to  the  right  from  U3  (at  (c))  as  shown,  and  the  distortion  of 
member  U2-L3  upward  and  parallel  to  U2-L3  from  L3  and  drawing 
perpendiculars  to  these  distortions,  we  obtain  point  U2,  which  shows  the 
position  of  joint  U2.  That  is,  joint  U2  has  moved  from  L3  to  U2.  Joint 
L2  moves  toward  U2  and  away  from  L3.  Then  laying  off  the  distortion 
of  member  U2-L2  upward  from  U2  and  the  distortion  of  L2-L3  to 
the  left  from  L3,  and  drawing  perpendiculars  to  these  distortions,  we 
obtain  point  L2,  which  shows  the  position  of  joint  L2.  Joint  Ul  moves 
toward  U2  and  away  from  L2.  Then  laying  off  the  distortion  of  member 
U1-U2  to  the  right  from  U2,  as  shown,  and  the  distortion  of  member  U1-L2 


Fig.   360 


DESIGN  OF  SIMPLE  KAILKOAD  BKIDGES 


527 


upward  and  parallel  to  U1-L2  from  L'2,  as  shown,  and  drawing  perpen- 
diculars to  these  distortions,  we  obtain  point  £71,  which  shows  the  loca- 
tion of  joint  £71.  Joint  LI  moves  downward  from  Ul  and  to  the  left  from 
L2.  So  laying  off  the  distortion  in  £71-L1  downward  from  £71  (at  (c)  ) 
and  the  distortion  of  L1-L2  to  the  left  from  L2  as  shown  and  drawing  per- 
pendiculars to  these  distortions  we  obtain  point  LI,  which  shows  the  loca- 
tion of  joint  LI.  Joint  LO  moves  toward  £71  and  away  from  LI.  Then 
laying  off  the  distortion  of  member  Ul-LO  upward  and  parallel  to  Ul-LO 
from  £71  and  the  distortion  of  member  LO-L1  to  the  left  from  LI,  and 
drawing  perpendiculars  to  these  distortions,  we  obtain  the  point  LO, 
which  shows  the  location  of  joint  LO.  We  now  have  the  position  of  the 
joints  of  the  left  half  of  the  truss  determined  at  (c)  in  reference  to  L3. 
As  is  obvious,  in  reference  to  that  point,  the  left  half  of  the  truss  would 
be  bent  upward,  as  indicated  by  the  dotted  outline,  and  of  course  the 
right  half  would  be  bent  up  the  same  amount.  Now,  it  is  evident  that 
the  deflection  of  the  bottom  chord  joints  is  the  distance  they  are  below  a 
horizontal  line  through  LO' .  These  distances  are  obtained  for  each  joint 
by  measuring  down  from  LO  (at  (c)).  We  thus  obtain  -£§" ,  J",  and 
l^V  for  the  deflection  of  joints  LI,  L2,  and  L3,  respectively.  Then  a 
diagram  of  these  deflections  can  be  drawn  as  shown  at  (b).  The  deflec- 


U2    -* 


b       U4 


L2 


L3 


(aj 


;&.  3d 


tion  of  any  ordinary  bridge  truss  can  be  determined  in  this  manner.  In 
case  of  odd  panels,  it  is  best  to  begin  at  the  middle  of  the  center  panel 
in  constructing  the  Williot  diagram.  For  example,  point  a  (Fig.  361  (a)) 
would  be  considered  fixed  in  position  and  line  ab  fixed  in  direction.  The 
diagonal  £73-L4  and  £74-L3  would  be  ignored,  as  they  would  have  no 
stress  in  them  —  assuming  that  a  uniform  live-  and  dead-load  extended 
over  the  entire  span.  The  point  b  would  move  downward  the  distance 
A2,  which  is  the  distortion  of  each  of  the  posts  £73-L3  and  £74-L4.  Then 
taking  a  as  the  starting  point  (at  (b)),  the  point  b  is  located  by  laying 
off  A2  downward  from  a,  as  shown.  Then  joint  L3  is  located  by  laying 
off  Al,  which  is  one-half  of  the  distortion  of  L3-L4,  to  the  left  from  a 
and  joint  £73  is  located  by  laving  off  A3,  which  is  one-half  of  the  distor- 
tion of  £73-£74,  to  the  right  "from  b.  Then,  having  joints  L3  and  £73 
located,  the  work  of  determining  the  deflection  proceeds  as  previously 
explained. 


528 


STRUCTURAL  ENGINEERING 


The  above  manner  of  determining  deflection  of  trusses  is  quite 
accurate  as  far  as  vertical  movement  is  concerned,  which  is  usually  all 
that  is  desired,  but  if  the  horizontal  movement  be  desired,  the  work  must 
be  carried  out  in  a  somewhat  different  manner.  It  will  be  seen  that  in 
the  above  case  we  assumed  the  center  of  the  truss  to  remain  fixed  and  that 
each  end  of  the  truss  was  free  to  move.  This  assumption  is  not  true,  as 
we  know,  as  one  end  of  the  truss  is  fixed  and  the  other  end  is  free  to  move. 


Fig.  362 


Now,  evidently  the  only  way  to  locate  correctly  the  deflected  posi- 
tion of  the  truss  would  be  to  begin  at  the  fixed  end. 

So  let  the  diagram  at  (a),  Fig.  362,  represent  the  same  truss  as 
considered  above,  and  suppose  the  end  L6  fixed  and  the  end  LO  free  (sup- 
ported upon  rollers).  Then  considering  LC>  fixed  in  position  and  member 
Z/6-L5  as  fixed  in  direction  and  taking  L6  (at  (b))  as  the  starting  point 


DESIGN  OF  SIMPLE  KAILROAD  BRIDGES  529 

and  constructing  the  Williot  diagram  as  shown,  we  obtain  the  relative 
position  of  the  joints,  but  the  truss  will  be  bent  upward  (so  to  speak)  as 
indicated  by  the  dotted  outline.  So,  to  obtain  the  correct  position  of  the 
joints,  the  truss  must  be  rotated  downward  about  L6,  until  the  joint  LO  is 
again  in  the  same  horizontal  line  as  L6.  This  means  that  joint  LO  will  be 
rotated  through  the  arc  ed.  This  arc  is  so  nearly  equal  to  the  vertical 
distance  A  that  it  can  be  considered  equal  to  it.  The  distance  A  is  equal 
to  the  vertical  distance  LO-LO' ' ,  shown  at  (b).  Now,  evidently,  all  of  the 
other  joints  of  the  truss  will  be  revolved  through  the  same  angle  as  LO 
and  the  arc  they  describe  in  each  case  will  be  directly  proportional  to 
their  radius.  These  arcs  are  so  small  that  the  distance  from  L6  to  the 
undeflected  position  of  any  joint  can  be  taken  as  its  radius.  Then  the 
radius  for  any  joint  of  the  bottom  chord  is  the  horizontal  distance  from 
L6  out  to  the  joint. .  For  example,  the  radius  for  joint  LI  is  L1-L6  ;  for 
L2  it  is  L2-L6;  and  so  on.  The  radius  for  joint  £71  is  J71-L6;  for  U2 
it  is  £72-L6 ;  and  so  on.  The  arcs  can  be  considered  as  straight  lines 
perpendicular  to  these  radii. 

Then  the  arcs  through  which  the  joints  LO,  LI,  L2,  L3,  L4,  and  L5 
are  rotated  are  represented,  respectively,  by  the  lines  A,  1,  2,  3,  4,  and 
5  and  the  arcs  through  which  joints  £75,  £74,  U3,  £72,  and  Ul  are  rotated 
are  represented,  respectively,  by  the  lines  6,  7/8,  9,  and  10.  Now,  having 
the  lengths  of  the  arcs  through  which  the  joints  are  rotated,  it  is  an 
easy  matter  to  determine  the  true  deflected  position  of  any  joint.  As  an 
example,  let  us  consider  joint  L4.  By  considering  the  member  L5-L6 
fixed  in  direction  the  joints  would  all  move  upward  to  the  positions  shown 
on  the  Williot  diagram  at  (b).  That  is,  each  would  move  from  L6  to 
the  position  indicated.  Then  laying  off  from  L4  downward  the  arc  4, 
we  obtain  the  point  L4',  which  is  the  true  deflected  position  of  joint  L4. 
That  is,  joint  L4  really  deflects  from  L6  to  L4'  and,  hence,  its  vertical 
movement  is  represented  by  the  distance  0-L4'  and  its  horizontal  move- 
ment by  the  distance  L6-O.  As  another  case,  let  us  consider  joint  £75. 
The  line  6,  which  is  perpendicular  to  the  end  post  £75-L6,  represents  the 
arc  through  which  this  joint  is  rotated.  So,  by  laying  off  this  arc  from 
£75  the  point  £75'  is  obtained,  which  is  the  true  deflected  position  of 
joint  £75.  That  is,  joint  £75  really  moves  from  L6  to  £75'.  The  true 
deflected  position  of  each  of  the  other  joints  can  be  determined  in  this 
manner,  but  time  can  be  saved  by  laying  off  all  of  the  arcs  upward  from 
L6.  In  applying  this  method  let  us  first  consider  joint  L4.  Laying  off 
arc  4  upward  from  L6,  we  obtain  point  L4".  Now,  the  deflected  posi- 
tion of  joint  L4  is  shown  by  the  line  L4"-L4.  This  is  readily  seen  to  be 
true  as  L6-L4"  is  the  arc  through  which  the  joint  is  rotated,  and  as  the 
joint  was  already  deflected  up  to  L4,  the  difference  between  the  two  posi- 
tions would  undoubtedly  be  the  actual  deflected  distance.  In  the  same 
manner,  laying  off  arc  7,  from  L6,  we  obtain  point  £74".  Then  the  actual 
movement  of  joint  £74  is  from  £74"  to  £74.  By  laying  off  all  the  arcs 
described  by  the  joints  from  L6  we  will  obtain  the  points  LO",  L\n ',  Ul" , 
£72",  etc.  If  these  be  connected  by  lines,  we  obtain  the  truss  shown,  which 
is  perpendicular  to  the  truss  shown  at  (a).  We  then  have  the  deflected  dis- 
tance of  each  joint  given.  For  example,  the  joint  LO  moves  from  LO"  to 
LO,  joint  LI  from  LI"  to  LI,  joint  £71  from  £71"  to  £71,  joint  £72  from 
£72"  to  £72,  and  so  on.  By  measuring  the  vertical  and  horizontal  com- 


530 


STEUCTUEAL  ENGINEEEING 


ponents  of  these  distances  the  true  deflected  position  of  each  joint  can  be 

plotted  as  shown  at  (c),  which  in  that  case  is  for  the  bottom  chord  joints. 

Now,  it  will  be  seen  that  all  we  need  do  to  determine  the  deflection 

of  the  truss  is  to  construct  the  Williot  diagram  at  (b)  and  draw  the  truss 

LQ"-U\" L6.     This  truss  can  be  drawn  by  dividing  the  line  LQ>-LO" 

into  as  many  equal  divisions  as  there  are  panels  in  the  bridge  and  then 
locating  one  of  the  points  £71" ',  £72" ',  etc.,  which  is  easily  done.  For  in- 
stance, £71"  is  located  by  drawing  from  LO"  a  line  perpendicular  to  end 
post  LO-U1  and  drawing  from  LI"  a  line  perpendicular  to  LG-LO" '.  The 
entire  truss  L0"-£71"  . .  .L6  of  course  can  be  constructed  in  the  .manner 
indicated  at  (b).  That  is,  by  projecting  points  -LI",  L2",  etc.,  over  from 
the  line  C-L6  and  then  locating  the  points  £71",  £72",  etc.,  by  laying  off 
from  LQ  the  arcs  10,  9,  etc. 

That  the  extremities  of  the  arcs  laid  off  from  LG  will  form  the  truss 
shown  at  (b)  is  readily  seen.  In  the  first  place,  there  is  no  question  as 
to  the  location  of  the  points  LO",  Z/l",  J/2"  .  .  .L6  and  it  is  seen  from 
the  construction  that  the  two  points  L4"  and  £74",  L3"  and  £73",  and 
so  on,  are  in  each  case  on  the  same  horizontal  line.  So  that  in  reality, 
the  only  thing  in  question  is  as  to  whether  the  points  £71",  £72",  £73",  etc., 
are  in  the  same  vertical  line.  This,  however,  is  readily  shown  to  be  true 
by  comparing  the  construction  at  (b)  with  the  diagram  at  (a).  The  arc 
L6-£74"  is  perpendicular  to  the  radius  L6-£74,  and,  as  is  seen  from  the 
construction  Z/4"  and  £74",  is  on  the  same  horizontal  line  and,  hence,  the 
triangles  L6-£74"-L4"  and  L6-£74-Z/4  are  similar,  the  corresponding  sides 
being  perpendicular  each  to  each.  In  the  same  manner  it  can  be  shown 
that  triangle  L6-£73"-L3"  is  similar  to  L6-£73-L3,  and  triangle  L6- 
£72"-L2"  is  similar  to  triangle  L6-U2-L2.  Then,  evidently,  as  £74-L4, 
£73-L3,  £72-L2,  etc.,  are  of  equal  length  at  (a),  the  lines  £74"-L4", 
£73"-L3"  will  be  of  equal  length  at  (b)  and,  hence,  the  points  £75",  £74", 
etc.,  will  be  in  the  same  vertical  line. 

The  determination  of  deflection  or  displacement  by  the  use  of  the 
Williot  diagram  and  the  rotation  diagram  combined,  as  shown  at  (b), 

was  first  proposed  by 
Professor  Mohr,*  and  the 
diagram  LO"-U\"  .  .  . L6 
is  known  as  Mohr's  rota- 
tion diagram." 

The  construction  of 
the  Mohr  rotation  dia- 
gram is  based  upon  the 
fact  that  the  lines  joining 
the  extremities  of  the  arcs 
described  by  the  joints  of 
a  truss  when  rotated 
through  a  small  angle 

about  any  point  will,  when  the  arcs  are  laid  off  from  a  point,  form  a  truss 
perpendicular  to  the  first  truss.  For  example,  suppose  the  truss  shown  at 
(a),  Fig.  363,  be  revolved  about  any  point  0  through  the  small  angle  6. 
Let  Al,  A2.  . . .A5  represent  the  arcs  described  by  the  joints.  Then,  by 


Fig.   363 


*  See  MoHtor's  Kinetic  Theory  of  Engineering  Structures. 


DESIGN  OF  SIMPLE  BAILEOAD  BRIDGES 


531 


laying  off  these  arcs  from  any  point  P,  as  shown  at  (b),  and  connecting 
their  extremities,  we  obtain  the  truss  shown  dotted.  This  truss  is  per- 
pendicular to  the  truss  at  (a).  All  of  this  is  readily  shown  to  be  true  by 
simple  geometrical  analysis. 

217.  Determination  of  Camber. — It  is  obvious  that,  if  each  com- 
pression member  in  a  truss  were  lengthened  an  amount  equal  to  its  dis- 
tortion and  each  tension  member  were  shortened  an  amount  equal  to  its 
distortion,  that  the  truss  when  supporting  no  load  would  be  cambered 
(curved  upward)  to  an  extent  equal  to  the  deflection  and  that  the  truss 
would  be  straight  when  supporting  the  maximum  full  load.  Camber  ob- 
tained in  this  manner  is  known  as  "exact  camber." 

In  the  case  of  trusses  up  to  300  ft.  in  length  sufficient  camber  seems 
to  be  obtained  by  merely  increasing  the  length  of  the  top  chord  about  ^ 
of  an  inch  for  each  10  feet  of  length,  as  explained  in  Art.  185,  but  longer 
spans  should  have  exact  camber. 

The  position  of  the  joints  due  to  camber,  in  the  case  of  exact  camber, 
is  determined  exactly  in  the  same  manner  as  the  deflection;  in  fact,  the 
movement  of  the  joints  is  exactly  equal  but  opposite  in  direction.  In  case 
the  camber  is  obtained  by  merely  lengthening  the  top  chord,  the  posi- 
tion of  the  joints  is  most  readily  obtained  by  the  graphical  method,  and 
the  work  is  practically  the  same  as  for  the  deflection.  As  an  example, 
let  it  be  required  to  locate  the  cambered  position  of  the  joints  of  the  truss 
shown  in  Fig.  279,  due  to  the  cambered  length  specified  for  that  struc- 
ture. (See  Art. '185.) 

First  draw  the  diagram  of  the  truss  shown  at  (a),  Fig.  364,  and 
write  on  each  of  the  members  the  amount  its  length  is  changed  to  obtain 
the  camber.  As  is  seen, 

only   the   lengths   of   the  ui    +1"    u?   +,i"  us  *-%"    U4  +£"    us 

top  chords  and  diagonals 
are  changed,  and  as  the 
lengths  are  increased  in 
every  case,  the  plus  sign 
is  used  throughout. 

We  draw  the  dia- 
grams at  (c)  by  taking 
L3  as  the  starting  point 
and  assuming  joint  Z/3 
as  fixed  in  position,  and 
member  C73-L3  fixed  in 
direction.  Joint  £73  does 
not  move  in  reference  to 
L3  (as  the  length  of  £73- 
Z/3  is  not  changed),  so 
£73  will  be  at  the  same 
point  as  Z/3.  Joint  £72  in 
reference  to  Z/3  moves  to 

the  left  -jffc-  in.  So  from  £73  (marked  Z/3  also)  lay  off  f$  in.  to  the  left, 
as  shown.  Joint  £72  in  reference  to  L3  moves  away  from  Z/3  the  dis- 
tance -fo  in.  So  from  £73  lay  off  ^\  *n-  upward  and  to  the  left  as  shown. 
Then  drawing  perpendiculars  to  these,  we  obtain  point  £72,  which  shows 
the  position  of  joint  £72.  Joint  Z/2  does  not  move  in  reference  to  either 


*'-'v^LZ 


NLO 

Fig.   364 


532  STEUCTUEAL  ENGINEEEING 

joint  U2  or  L3.  So  drawing  from  point  U2  a  perpendicular  to  member 
U2-L2  and  from  L3  a  perpendicular  to  member  L3-L2,  we  obtain  point 
L2,  which  shows  the  location  of  joint  L2.  To  aid  in  understanding  this 
step,  we  can  imagine  the  change  in  length  of  each  of  the  members 
U2-L2  and  L2-L3  as  being  infinitesimal  and  the  lines  U2-L2  and  L2-L3 
as  being  perpendicular,,  respectively,  to  these  changes. 

Joint  £71  moves  -f^  in.  away  from  U2  and  -/%  in.  from  L2.  Then 
from  U2  (at  (c))  lay  off  -f^  in.  to  the  left  and  from  L2  lay  off  ^  in' 
upward  and  to  the  left,  and  then  drawing  perpendiculars  we  obtain  point 
£71,  which  shows  the  position  of  joint  Ul.  Joint  LI  does  not  move  in 
reference  to  either  joint  £71  or  L2.  Then  by  drawing  from  £71  a  line 
perpendicular  to  member  £71-L1  and  from  L2  a  line  perpendicular  to 
member  L1-L2,  we  obtain  the  point  LI,  which  shows  the  location  of  joint 
LI.  Likewise,  joint  LO  does  not  move  in  reference  to  either  joint  £71 
or  LI.  Then  by  drawing  from  £71  a  line  perpendicular  to  member  £71-LO 
and  from  LI  a  line  perpendicular  to  member  LO-L1,  we  obtain  point 
LO,  which  shows  the  location  of  joint  LO.  Now,  we  have  the  position  of 
the  joints  shown  at  (c)  for  the  left  half  of  the  truss,  and  as  the  truss 
is  symmetrical  about  the  center  of  the  span  the  diagram  at  (c)  really 
shows  the  position  of  all  the  joints  in  the  truss.  The  diagram  at  (b) 
shows  the  positions  of  the  joints  of  the  lower  chord  which  are  obtained 
from  the  diagram  at  (c)  by  measuring  upward  from  LO. 

It  is  interesting  to  note  how  near  the  camber  shown  at  (b)  comes 
to  being  equal  to  the  deflection  found  for  this  same  truss  in  Art.  215. 

The  cambered  position  of  the  joints,  especially  the  lower  chord 
joints,  of  long  spans  is  required  mostly  for  locating  the  tops  of  the  camber 
blocks  upon  which  the  truss  is  supported  while  the  structure  is  being 
erected.  These  supports  or  blocks,  placed  under  each  panel  point,  should 
have  about  the  correct  height,  for  otherwise  the  truss  members  will  not 
fit  into  place. 


CHAPTER  XII 

DESIGN  OF  SIMPLE  HIGHWAY  BRIDGES 

218.  Types. — Steel  highway  bridges  can  be  divided  into  the  fol- 
lowing types :   beam  bridges,  plate  girders,  viaducts,  pony  truss,  and  high 
truss  bridges.     Beam  bridges  and  viaducts  are  usually  deck  structures, 
while  pony  truss  bridges  are  always  through  structures.     Plate  girders 
and  high  truss  bridges  may  be  either  through  or  deck  structures. 

219.  Live  Load. — The  maximum  live  load  depends,  as  a  rule,  upon 
the  locality  of  the  structure.     It  varies  from  street  cars,  heavy  traction 
engines,  trucks,  and  dense  crowds  of  people  in  and  near  cities  and  towns, 
to  light  traction  engines  and  droves  of  live  stock  in  the  country  districts. 
So,  in  respect  to  the  live  load,  highway  bridges  can  be  classed,  except  for 
isolated  cases,  as  city  bridges  and  country  bridges.     As  each  city  bridge, 
as  a  rule,  is  a  special  problem,  we  shall  consider  only  country  bridges, 
although  the  general  principles  involved  in  the  design  of  the  two  classes 
are  identical. 

The  live  load  for  country  bridges  is  specified  as  a  uniform  load  of 
so  many  pounds  per  square  foot  of  floor — the  longer  the  span,  the  less 
the  load — or  a  load  of  so  many  tons  concentrated  on  two  axles  from  10 
to  12  ft.  apart,  the  transverse  distance  between  the  centers  of  the  wheels 
being  from  5  to  7  ft.  The  concentrated  load  is  used  where  it  produces 
a  greater  stress  than  the  uniform  load.  The  intensities  of  the  loads  used 
vary  for  different  localities  depending,  of  course,  in  each  case,  upon  the 
actual  loads  found  to  exist  and  upon  the  probable  future  loads.  As  a  rule, 
the  influence  of  the  latter  would,  as  is  obvious,  depend  upon  the  financial 
status  of  the  locality  paying  for  the  structure.  The  live  loads  that  the 
author  considers  satisfactory  for  most  country  bridges  are  specified  farther 
on  in  the  "Specifications  for  Steel  Highway  Bridges." 

220.  Impact. — From  actual  tests  made  on  highway  bridges  over  the 
State  of  Iowa,  under  the  joint  auspices  of  the  State  Highway  Commis- 
sion and  the  Iowa  State  College  Engineering  Experiment  Station,  it  was 
found  that  impact  due  to  maximum  loads  is  negligible.     In  some  cases  the 
impact   due   to   some   light   loads — for   instance,   a   trotting   horse — was 
found  to  be  quite  high,  especially  for  bridges  having  wood  floors,  but  in 
all  such  cases  the  stresses  were  so  low  that  the  impact  in  no  way  could 
influence  the  design  of  such  structures.     In  the  case  of  bridges  having 
concrete  floors  no  impact  of  any  consequence  was  found  in  any  case. 

These  tests  were  conducted  by  the  author,  assisted  by  Mr.  C.  S. 
Nichols,  Assistant  Director  of  the  Experiment  Station,  and  others.  The 
results  were  verified  by  tests  made  in  the  State  of  Illinois  by  Prof. 
F.  O.  Dufour  and  State  Engineer  A.  N.  Johnson. 

Judging  from  the  above,  no  impact  should  be  considered  in  designing 
highway  bridges. 

533 


534  STRUCTURAL  ENGINEERING 

221.  Dead  Load.  —  The  dead  load,  as  in  the  case  of  railroad 
bridges,  consists  of  the  weight  of  metal  in  the  structure  and  the  weight 
of  the  floor. 

Owing  to  the  wide  variation  of  the  weight  of  metal  in  highway 
bridges  due  to  the  design,  details,  width  of  roadway,  loading,  and  so  on, 
it  is  practically  impossible  to  give  formulas  for  the  weight  of  metal  in 
general. 

In  the  case  of  beam  and  plate  girder  bridges,  the  weights  of  metal 
in  the  different  parts  are  readily  assumed  and  verified  as  the  calculations 
for  the  designs  are  made.  The  approximate  weight  of  metal  in  the 
structures  designated  can  be  obtained  from  the  following  formulas  where 
p  =  weight  of  metal  in  pounds  per  foot  of  bridge,  and  L  —  length  of  span 
in  feet,  c.c.  end  bearings  : 

For  pony  trusses  with  concrete  floors  (without  joists) 


(1). 
For  through  trusses  with  concrete  floors  (with  joists) 

(2). 


The  above  formulas  are  for  16-ft.  roadways.  To  obtain  the  weight 
of  metal  for  wider  roadways  add  15  Ibs.  for  each  additional  foot  of  width. 

The  weights  obtained  from  the  above  formulas  will,  as  a  rule,  be 
within  10  per  cent  of  the  actual  weight.  The  weight  of  reinforcing  steel 
in  the  concrete  floors  is  not  included.  The  weight  of  the  floor  in  each 
case  must  be  computed  and  added  to  the  weight  of  metal. 

222.  Specifications.  —  Quite  a  number  of  very  good  specifications 
for  steel  highway  bridges  have  been  written  and  published  of  late  years, 
but  unfortunately  these  do  not  agree  on  some  important  points  and  for 
that  reason  none  of  them  will  be  designated  here  as  a  standard,  but  instead 
the  following  specifications  by  the  author,  mostly  a  compilation,  will  be 
used  in  this  work: 


SPECIFICATIONS  FOR  STEEL  HIGHWAY  BRIDGES 
TYPES  OF  BRIDGES 

1.  The  following  types  are  recommended : 

Spans  up  to     30  feet — Rolled  I-beams, 
Spans  30  to     50  feet — Plate  girders, 
Spans  40  to     90  feet — Pony  trusses, 

Spans  90  to  150  feet — Riveted  high  trusses, 

Spans    over  150  feet — Pin-connected  trusses. 

LOADING 

2.  Dead  Load  shall  consist  of  the  entire  weight  of  structure,  including  the 
metal,  concrete  floor  or  wood  floor  (when  allowed),  and  the  fill  on  concrete  floors. 

In  estimating  the  dead  load  concrete  shall  be  considered  as  weighing  150 
Ibs.  and  fill  120  Ibs.  per  cu.  ft.,  and  timber  as  4.25  Ibs.  per  ft.  B.  M. 

The  dead  load  used  in  figuring  the  dead-load  stresses  must  be  within  10  per 
cent  of  the  actual  dead  weight. 


DESIGN  OF  SIMPLE  HIGHWAY  BRIDGES 


535 


3.  The  Live  Load  shall  be  as  follows : 

The  live  load  on  the  roadway  shall  be  taken  as  a  uniform  load  of 

100  Ibs.  per  sq.  ft.  for  all  spans  up  to  50  ft., 

90  Ibs.  per  sq.  ft.  for  spans      50   to    100   ft., 

80  Ibs.  per  sq.  ft.  for  spans    100   to    150    ft., 

70  Ibs.  per  sq.  ft.  for  spans    150   to    200   ft., 

60  Ibs.  per  sq.  ft.  for  spans  over  200  ft., 

or  a  15-ton  engine,  as  per  Fig.  365,  using  the  engine  whenever  greater  stresses  are 
obtained  than  by  using  the  uniform  load. 

The  live  load  on  sidewalks  shall  be  taken  as  a  uniform  load  of  50  Ibs.  per 
sq.  ft. 

4.  Wind  Load  for  high  truss  bridges  shall  be  300  Ibs.  per  lineal  ft.  of  span 
on  the  loaded  chord  and  100  Ibs.  per  lineal  ft.  on  the  unloaded  chord.     The  wind 


II- 0 


Fig.  365 


Fig.  366 


load  on  all  other  structures  shall  be  taken  as  30  Ibs.  per  sq.  ft.  on  one  and  one- 
half  times  the  vertical  projection  of  the  structure,  all  of  these  loads  to  be  consid- 
ered as  moving  loads. 

CLEARANCE 

5.  The  Clearance  shall  not  be  less  than  shown  on  the  diagram  in  Fig.  366. 

MATERIAL 

6.  All  material  shall  be  of  medium  steel  except  rivets,  bolts,  rockers,  shoes, 
and  pedestals.     Rivets   and  bolts  shall  be  of  rivet   steel.     Kockers,   shoes,  and 
pedestals  shall  be  of  cast  iron  or  cast  steel. 

7.  Medium  Steel  shall  have  an  ultimate  strength  of  60,000  to  70,000  Ibs. 
per  sq.  in. ;  an  elastic  limit  of  not  less  than  one-half  the  ultimate  strength ;  elonga- 
tion twenty-two  per  cent  (22%}  ;  bending  test  through  180  degrees  to  a  diameter 
equal  to  the  thickness  of  the  piece  tested  without  fracture  on  outside  of  bent  por- 
tion.    It  shall  not  contain  more  than  0.05  per  cent  of  sulphur  and  if  basic  shall 
not  contain  more  than  0.04  per  cent  of  phosphorus  or  0.06  per  cent  if  acid. 

8.  Eivet  Steel  shall  have  an  ultimate  strength  of  48,000  to  58,000  Ibs.  per 
sq.  in. ;  elastic  limit  of  not  less  than  one-half  of  the  ultimate  strength ;  elongation 
twenty-six  per  cent  (26%)  ;  bending  test  180  degrees  flat  on  itself  without  fracture 
on  outside  of  bent  portion.    It  shall  not  contain  more  than  0.04  per  cent  of  either 
sulphur  or  phosphorus. 

9.  Cast  Steel  shall  have  an  ultimate  strength  of  not  less  than  65,000  Ibs.  per 
sq.  in.;  elongation  of  fifteen  per  cent  (15%)  ;  bending  test  through  90  degrees  to 
a  diameter  of  three  times  the  thickness  of  the  piece  tested  without  fracture  on 
outside  of  bent  portion.     It  shall  not  contain  more  than  0.08  per  cent  phosphorus 
nor  more  than  0.05  per  cent  of  sulphur. 

10.  All  of  the  above  steel  shall  be  made  by  the  open  hearth  process. 

11.  Cast  Iron  shall  be  what  is  known  as  grey  iron.     A  test  piece  one  inch 
square  in  cross-section,  loaded  in  the  middle  between  supports  12  ins.  apart,  shall 
support  at  least  2,500  Ibs.  and  deflect  at  least  0.15  in.  before  rupture. 


536  STEUCTUEAL  ENGINEEEING 

ALLOWABLE  UNIT  STRESSES 

12.  The  Unit  Stress  in  any  part  of  any  structure  due  to  dead  and  live  load 
combined  shall  not  exceed  the  following: 

For  Metal 

Axial  tension  on  net  section 16,000  Ibs.  per  sq.  in. 

Axial  compression  on  gross  section  where  L  is  length  of 

member  in  ins.  and  r  the  least  radius  of  gyration 

in  ins 16,000  -  70  (L/r)   Ibs.  per  sq.  in. 

Shear  on  shop  rivets  and  pins 12,000  Ibs.  per  sq.  in. 

Shear  on  field  rivets,  webs  of  girders  and  rolled  beams.  .  .  .  10,000  Ibs.  per  sq.  in. 

Shear  on  bolts 9,000  Ibs.  per  sq.  in. 

Bearing  on  shop  rivets  and  pins 24,000  Ibs.  per  sq.  in. 

Bearing  on  field  rivets 20,000  Ibs.  per  sq.  in. 

Bearing  on  bolts 18,000  Ibs.  per  sq.  in. 

Bending  on  pins,  rivets  and  bolts ._.  25,000  Ibs.  per  sq.  in. 

Bending  on  beams  and  girders ".  16,000  Ibs.  per  sq.  in. 

Bearing  on  rockers  or  rollers  where  d  is  the  diameter  of 

rocker  or  roller  in  ins (600  X  d)  Ibs.  per  lin.  in. 

Pin-bearing  on  rockers 12,000  Ibs.  per  sq.  in. 

For  Timber 
(Douglas  fir,  white  oak,  and  long  leaf  yellow  pine) 

Bending   1,500  Ibs.  per  sq.  in. 

Tension  with  grain 1,500  Ibs.  per  sq.  in. 

Compression  with  grain 1,500  Ibs.  per  sq.  in. 

Shear  across  grain 800  Ibs.  per  sq.  in. 

Concrete 

Compression  due  to  cross  bending 700  Ibs.  per  sq.  in. 

Direct  compression  and  bearing  of  shoes  and  pedestals  on 

masonry 600  Ibs.  per  sq.  in. 

Tension  000  Ibs.  per  sq.  in. 

PROPORTION  OF  PARTS 

13.  The  lengths  of  compression  members  shall  not  exceed  120  times  their 
least  radius  of  gyration  in  case  of  pin-connected  members  nor  more  than  125  times 
in  case  of  members  having  riveted  end  connections. 

14.  Members  subjected  to  alternate  stresses  of  tension  and  compression  shall 
be  proportioned  for  each  kind  of  stress. 

15.  Whenever  the  live-  and  dead-load  stresses  are  of  opposite  signs  only  two- 
thirds  of  the  dead-load  stress  shall  be  considered  as  counteracting  the  live-load 
stress. 

16.  Members  subjected  to  both  axial  and  bending  stresses  shall  be  propor- 
tioned so   that  the  combined  stresses  will  not  exceed  the  allowed  axial  stress, 
although  if  the  bending  is  due  to  wind  load  the  axial  stress  can  be  increased  25 
per  cent  over  the  allowable  specified  above. 

17.  In  proportioning  tension  members,  the  net  area  of  cross-section  shall 
be  considered  as  the  effective  section  and  in  deducting  for  rivet  holes  the  diameter 
of  each  hole  shall  be  considered  as  being  &  in.  larger  than  the  nominal  diameter  of 
the  rivet. 

18.  In  determining  the  number  of  rivets   the  nominal  diameter  shall  be 
used. 

19.  Pin-connected  riveted  tension  members  shall  have  a  net  section  through 
the  pin  hole  at  least  25  per  cent  in  excess  of  the  net  section  of  the  body  of  the 
member,  and  the  net  section  back  of  the  pin  hole,  parallel  with  the  axis  of  the 
member,  shall  be  not  less  than  the  net  section  of  the  body  of  the  member. 

20.  Plate  girders  shall  be  proportioned  either  by  the  moment  of  inertia  of 
their  net  section,  or  by  assuming  that  the  areas  of  the  flanges  are  concentrated 
at  their  centers  of  gravity,  in  which  case  one-eighth  of  the  gross  section  of  the 
web,  if  properly  spliced,  may  be  used  as  flange  section. 

21.  The  gross  section  of  compression  flanges  of  plate  girders  shall  be  not 
less  than  the  gross  section  of  the  tension  flange  and  the  unsupported  distance  of 
any  compression  flange  shall  be  not  greater  than  15  times  the  width  of  the  flange. 


DESIGN  OF  SIMPLE  HIGHWAY  BRIDGES  537 

22.  The  flanges  of  plate  girders  shall  be  connected  to  the  web  with  a  suf- 
ficient number  of  rivets  to  transmit  the  flange  increment  combined  with  any  load 
applied  directly  on  the  flange  and,  in  case  cover  plates  are  used,  the  cross-section 
of  these  plates  must  not  exceed  the  cross-section  of  the  flange  angles. 

DETAILS  OF  DESIGN 

23.  The  strength  of  end  connections  shall  be  sufficient  to  develop  the  full 
strength  of  the  member,  even  though  the  computed  stress  is  less,  the  kind  of  stress 
to  which  the  member  is  subjected  being  considered. 

24.  The  minimum  thickness  of  metal  shall  be  £  in.  except  for  fillers  and 
webs  of  channels. 

The  minimum  distance  between  centers  of  rivet  holes  shall  be  three  diameters 
of  the  rivet;  but  preferably  the  distance  shall  be  not  less  than  3  ins.  for  |-in. 
rivets,  2£  ins.  for  f-in.  rivets,  and  2  ins.  for  f-in.  rivets.  The  maximum  pitch  shall 
be  6  ins.  for  f-in.  rivets,  5  ins.  for  f-in.  rivets,  and  4£  ins.  for  f-in.  and  *-in.  rivets. 
Where  two  or  more  plates  are  used  in  contact,  rivets  not  more  than  12  ins.  apart 
in  either  direction  shall  be  considered  sufficient  to  hold  the  plates  well  together. 

25.  The  minimum  distance  from  the  center  of  any  rivet  hole  to  a  sheared 
edge  shall  be  1^  ins.  for  |-in.  rivets,  11  ins.  for  f-in.  rivets,  and  1£  ins.  for  f-in. 
and  i-in.  rivets.     The  maximum  distance  from  any  edge  shall  be  eight  times  the 
thickness  of  the  plate,  but  shall  not  exceed  6  ins.  in  any  case. 

26.  The  pitch  of  rivets  at  the  ends  of  built  compression  members  shall  not 
exceed  four  diameters  for  a  distance  equal  to  one  and  one-half  times  the  maximum 
width  of  the  member.     In  compression  members  the  metal  shall  be  concentrated 
as  much  as  possible  in  webs  and  flanges.     The  thickness  of  each  web  shall  not  be 
less  than  one-fortieth  of  the  distance  between  its  connections  to  the  flanges.    Cover 
plates  shall  have  a  thickness  not  less  than  one-fiftieth  of  the  distance  between 
rivet  lines. 

27.  Flanges  of  girders  without  cover  plates  shall  have  a  minimum  thickness 
of  one-twelfth  of  the  width  of  the  outstanding  legs. 

28.  The  open  sides  of  compression  members  shall  be  provided  with  lattice 
and  shall  have  tie  plates  as  near  each  end  as  practicable.     Tie  plates  shall  be 
provided  at  intermediate  points  where  the  lattice  is  interrupted.    In  main  members 
the  end  tie  plates  shall  have  a  length  not  less  than  their  width  and  intermediate 
ones  not  less  than  one-half  their  width.     Tie  plates  shall  have  a  thickness  of  not 
less  than  one-fiftieth  of  the  distance  between  rivet  lines. 

29.  The  minimum  width  of  lattice  bars  shall  not  be  less  than  2£  ins.  for 
f-in.  rivets,  2£  ins.  for  f-in.  rivets,  2  ins.  for  f-in.  rivets,  and  If  ins.  for  ^-in. 
rivets.     The  thickness  of  lattice  bars  shall  not  be  less  than  one-fiftieth  of  the  dis- 
tance between  end  rivets  for  single  lattice  and  one-sixtieth  for  double  lattice. 

30.  The  inclination  of  lattice  bars  with  the  axis  of  the  member  shall  be 
not  less  than  45  degrees,  and  when  the  distance  between  rivet  lines  in  the  flanges  is 
more  than  15  ins.,  double  lattice  shall  be  used  which  shall  be  riveted  at  inter- 
sections. 

31.  Abutting  joints  in  top  chords  of  high  trusses  when  faced  for  bearing 
shall  be  spliced  on  four  sides  sufficiently  to  hold  the  connecting  members  accurately 
in  place.    All  other  joints  in  riveted  work,  whether  in  tension  or  compression,  shall 
be  fully  spliced. 

32.  Pin-holes  shall  be  reinforced  by  plates  where  necessary,  and  at  least  one 
plate  shall  be  as  wide  as  the  flanges  will  allow  and  be  on  the  same  side  as  the 
flange  angles.    They  shall  contain  sufficient  rivets  to  distribute  their  portion  of  the 
pin  pressure  to  the  full  cross-section  of  the  member. 

33.  In  case  special  permission  be  given,  the  field  connections  of  small  bridges 
may  be  bolted.    This  refers  to  the  connections  of  laterals,  transverse  bracing,  joists, 
floor  beams,  hand  rails,  and  main  splices  in  trusses.     The  holes  in  the  connections 
of  the  floor  beams  and  joists    (when  the  joists  connect  directly  to  the  web  of 
the  floor  beams)  must,  if  bolted,  be  sub-punched  and  reamed  to  iron  templets  and 
the  open  holes  in  the  main  splices  of  trusses  must  be  sub-punched  and  reamed  to 
size  at  the  shop  while  the  trusses  are  assembled.    The  bolts  used  must  have  hexa- 
gonal heads  and  nuts  and  standard  washers  and  must  fit  tightly  into  the  holes. 
If  a  desired  fit  can  not  be  obtained  without,  turned  bolts  must  be  used.     In  no 
case  will  the  bolting  of  the  connections  of  trusses  shipped  " knocked  down"  be 
permitted.    Eivets  are  preferable  to  bolts. 


538 


STRUCTURAL  ENGINEERING 


34.  Rivets  carrying  stress  and  passing  through  fillers  shall  be  increased  50 
per  cent  in  number;  and  the  excess  rivets,  where  possible,  shall  be  outside  of  the 
connected  member. 

35.  Provision  for  expansion  and  contraction  to  the  extent  of  £  in.  for  each 
10  ft.  of  length  shall  be  made  for  all  bridges. 

Spans  under  70  ft.  in  length  shall  be  fixed  at  one  end  and  allowed  to  expand 
and  contract  upon  planed  surfaces  at  the  other  end.  Both  ends  shall  be  anchored. 
Slotted  holes  for  the  anchor  bolts  shall  be  provided  at  the  sliding  end. 

The  masonry  plates  or  pedestals  shall  be  cast  iron  in  all  cases. 

36.  The  ends  of  65-ft.  spans  and  over  shall  be  pin-bearing  and  have  cast 
rockers  at  one  end  and  fixed  cast  shoes  at  the  other.     The  rockers  shall  be  of  the 
style  shown  in  Fig.  367. 

The  fixed  shoes  shall  be  securely  anchored  to  the  masonry  and  the  rockers 
shall  be  anchored  as  indicated  in  Fig.  367. 


Fig.  367 

37.  All  laterals  and  transverse  bracing  in  all  structures  shall  be  composed 
of  rigid  members. 

38.  The  webs  of  plate  girders  shall  be  stiffened  by  stiffening  angles  gen- 
erally in  pairs  over  bearings  at  points  of  concentrated  loads  and  at  other  points 
where  the  thickness  of  the  web  is  less  than  one-sixtieth  of  the  unsupported  distance 
between  flange  angles.    The  distance  between  stiff eners  shall  not  exceed  that  given 
by  the  following  formula,  with  a  maximum  limit  of  6  ft.,  but  in  no  case  exceeding 
the  depth  of  the  girder: 

d=±   (12,000-*) 

where  d  =  clear  distance  between  stiffeners  or  flange  angles, 
t  =  thickness  of  web, 
s  =  shear  per  square  inch. 

39.  The  stiffeners  at  ends  and  at  points  of  concentrated  loads  shall  be  pro- 
portioned by  the  formula  of  Paragraph  12 — (16,000-70    (L/r)  ) — the  effective 
length  being  taken  as  one-half  the  depth  of  the  girder.    End  stiffeners  and  those 
under  concentrated  loads  shall  be  on  fillers  and  have  their  outstanding  legs  as  wide 
as  the  flange  angles  will  allow.     Intermediate  stiffeners  may  be  on  fillers  or  be 
crimped  and  their  outstanding  legs  shall  be  not  less  than  one-thirtieth  of  the  depth 
of  the  girder  plus  2  ins.    All  stiffeners  shall  fit  tightly  against  the  flange  angles. 

40.  Through  plate  girders  shall  have  their  top  flanges  stayed  at  each  end 
of  every  floor  beam,  or  in  the  case  of  solid  floors  at  a  distance  not  to  exceed  12  ft., 
by  brackets  or  gusset  plates. 

41.  Pony  trusses  shall  be  riveted  structures,  with  double-webbed  chords  and 
shall  have  web  members  latticed  or  otherwise  effectively  stiffened. 

42.  The  top  chords  of  pony  trusses  under  70  ft.  in  length  shall  be  securely 
held  in  position  at  the  panel  points  by  gusset  plates,  knee  braces,  or  solid  webbed 
vertical  posts  connected  rigidly  to  the  floor  beams.    The  top  chords  of  pony  trusses 
70  ft.  in  length  and  over  shall  be  stiffened  by  knee  braces  in  addition  to  having 
solid  webbed  posts  rigidly  connected  to  the  floor  beams. 


DESIGN  OF  SIMPLE  HIGHWAY  BKIDGES  539 

43.  Trusses  supporting  wooden  floors  shall  be  given  a  camber  by  increasing 
the  length  of  the  top  chord  J  in.  for  each  10  ft.  of  length  and  trusses  supporting 
concrete  floors  shall  be  given  a  camber  by  increasing  the  length  of  the  top  chord 
&  in.  for  each  10  ft.  of  length. 

Hip  verticals  and  similar  members  and  the  two  end  panels  of  the  bottom 
chords  of  pin-connected  bridges  shall  be  rigid  members. 

44.  All  web  splices  in  plate  girders  shall  be  sufficient  to  transmit  the  shear 
and  bending  on  the  web  at  the  point  of  splice. 

45.  The  eye-bars  composing  a  member  shall  be  so  arranged  that  adjacent 
bars  shall  not  have  their  surfaces  in  contact;  they  shall  be  as  nearly  parallel  to 
the  axis  of  the  truss  as  possible.     The  maximum  inclination  of  any  bar  is  limited 
to  1  in.  in  16  ft.,  and  the  bars  composing  a  member  must  all  have  about  the  same 
inclination. 

WORKMANSHIP 

46.  All  parts  forming  a  structure  shall  be  built  in  accordance  with  approved 
drawings.     The  workmanship  shall  be  equal  to  the  best  practice  in  modern  bridge 
works. 

47.  Material  shall  be  thoroughly  straightened  in  the  shop,  by  methods  that 
will  not  injure  it  before  being  laid  off  or  worked  in  any  way. 

48.  The  size  of  rivets  called  for  on  the  plans  shall  be  understood  to  mean 
the  actual  size  of  the  cold  rivet. 

49.  The  diameter  of  the  punch  shall  not  be  more  than  ^  in.  greater  than 
the  diameter  of  the  rivet ;  nor  the  diameter  of  the  die  more  than  &  in.  greater  than 
the  diameter  of  the  punch. 

50.  Punching  shall  be  accurately  done.     Drifting  to  enlarge  unfair  holes 
will  not  be  allowed.    If  the  holes  must  be  enlarged  to  admit  the  rivet  they  shall  be 
reamed.    Poor  matching  of  holes  will  be  cause  for  rejection. 

51.  Where  sub-punching  and  reaming  are  required,  the  punch  used  shall 
have  a  diameter  of  &  in.  smaller  than  the  nominal  diameter  of  the  rivet.     The 
holes  shall  then  be  reamed  to  a  diameter  ^  in.  larger  than  the  nominal  diameter 
of  the  rivet  used. 

If  necessary  to  take  the  pieces  apart  for  shipping  and  handling,  the  re- 
spective pieces  reamed  together  shall  be  match-marked  so  that  they  may  be  re- 
assembled in  the  same  position  in  the  final  setting  up.  That  is,  no  interchanging 
of  reamed  parts  will  be  permitted. 

52.  Eiveted  members  shall  have  all  parts  well  assembled  and  firmly  drawn 
together  with  bolts  before  riveting  is  commenced. 

53.  Web  plates  of  girders,  which  have  no  cover  plates,  shall  be  flush  with 
the  backs  of  the  flange  angles. 

54.  Eivets  shall  look  neat  and  finished  with  heads  of  approved  shape,  full 
and  of  equal  size.     They  shall  be  central  on  shank  and  grip  the  assembled  pieces 
firmly.     Eecupping  and  calking  will  not  be  allowed.     Loose,  burned,  or  otherwise 
defective  rivets   shall  be  cut   out  and  replaced.     All  rivets  shall  be  driven  by 
pressure  tools  wherever  possible.    Pneumatic  hammers  shall  be  used  in  preference 
to  hand-driving. 

55.  Eye-bars  shall  be  straight  and  true  to  size  and  shall  be  free  from  twists, 
folds  in  the  neck  or  head,  or  any  other  defects.     The  heads  shall  be  made  by  up- 
setting, rolling,  or  forging.     Welding  will  not  be  allowed.     The  heads  shall  be  so 
proportioned  that  the  bars  will  break  in  the  body  when  tested  to  rupture. 

•  56.  Before  boring  the  pin-holes  each  eye-bar  shall  be  properly  annealed  and 
carefully  straightened.  The  pin-holes  shall  be  in  the  center  line  of  the  bars  and 
in  the  center  of  the  heads.  Bars  of  the  same  length  shall  be  bored  so  accurately 
that,  when  placed  together,  pins  jfe  in.  smaller  in  diameter  than  the  pin-holes  can 
be  passed  through  the  holes  at  both  ends  of  the  bars  at  the  same  time  without 
forcing. 

57.  The  distance  center  to  center  of  pin-holes  shall  be  correct  within  -^ 
in.,  and  the  diameter  of  the  holes  not  more  than  ^  in.  larger  than  that  of  the  pin, 
for  pins  up  to  5  ins.  in  diameter,  and  3*3  in.  for  larger  pins. 

58.  Pins  shall  be  accurately  turned  and  shall  be  straight  and  smooth  and 
entirely  free  from  flaws. 


540 


STRUCTURAL  ENGINEERING 


PAINTING 

59.  All  metal  work  before  leaving  the  shops  shall  be  thoroughly  cleaned  from 
all  loose  scale  and  rust  and  be  given  one  coat  of  pure,  boiled  linseed  oil,  well 
worked  into  all  joints  and  open  spaces,  except  pin-holes  and  finished  surfaces, 
which  shall  be  given  one  coat  of  white  lead  and  tallow. 

All  surfaces  of  riveted  work  coming  into  contact  shall  be  painted  before 
being  riveted  together  with  one  good  coat  of  pure,  boiled  linseed  oil  and  red  lead. 

After  the  structure  is  erected  the  metal  work  shall  be  thoroughly  and  evenly 
painted  (cleaning  parts  when  necessary)  with  two  coats  of  either  red  lead  and 
linseed  oil  or  sublimed  blue  lead  and  linseed  oil,  tinted  as  directed. 


BEAM  BRIDGES 
Complete  Design  of  an  18-Ft.  Beam  Span  with  Wood  Floor 

223.    Data.— 

Length  of  span  =  18'0"  c.c.  of  end  bearings, 

Roadway  =  16'0"  clear, 

Live  load,  15-ton  traction  engine  as  per  Fig.  365,  Art.  222. 

Dead  load  as  hereafter  determined. 

224.    Design  of  Floor. — The  beams  will 


to" 


ii 

^~^~^_ 

3'^ 

» 

/ 

\ 

y 

a         b 

ti 

~7< 

,3' 

7" 

27" 

«* 


(b) 
Fig.  368 


be  spaced  about  27"  center  to  center.  The 
structure  must  be  designed  so  that  planks  16' 
long  can  be  used,  as  odd  lengths  are  usually  ex- 
pensive and  difficult  to  obtain.  Let  us  assume 
that  3"xl2"  yellow  pine  plank  will  be  used. 
The  large  wheels  on  a  traction  engine  are 
about  20"  wide.  Owing  to  the  plank  deflect- 
ing, the  wheel  of  a  traction  engine  will  bear 
mostly  at  its  edges  and  consequently  the  pres- 
sure on  the  plank  will  be  somewhat  as  indi- 
cated at  (o),  Fig.  368.  From  this  sketch  it  is 
seen  that  we  can  justly  assume  the  load  on 
the  wheel  to  be  applied  at  two  points,  a  and  b, 
one-half  of  the  total  load  at  each  point.  As- 
suming the  plank  as  forming  a  series  of  fixed 
beams,  we  have  at  (a)  the  case  of  a  fixed  beam 
supporting  two  equal  loads  of  5,000  #  each. 
In  applying  the  formulas  of  Art.  69,  L  equals 
27",  and  with  reference  to  support  A,  kL 
equals  7"  for  the  load  at  a  and  20"  for  the 
load  at  b.  Then  we  have  k  equals  -fa  in  the 
first  case  and  -|  ?  in  the  second  case. 

Now  applying  (7)  of  Art.  69,  we  have 


AT,  5,000  x2-[2  (1)'  _  (i)'-i 


for  the  moment  at  A.     Substituting  in  a  similar  manner  in  (6),  Art.  69, 

•we  obtain  5,000  for  the  shear  at  A.     Now,  having  the  bending  moment 

and  shear  at  A   determined,  the  bending  moment  on  the  plank  at  any 


DESIGN  OF  SIMPLE  HIGHWAY  BEIDGES  541 

point  between  the  beams  is  readily  determined.  The  maximum  will  be 
at  the  loads,  being  the  same  at  each  load.  Then  we  have 

M  =  -25,920  +  5,000x7  =  9,080  inch  Ibs. 

for  the  moment  at  a.  In  the  case  of  heavy  wagons  the  loading  on  the 
plank  would  be  as  shown  at  (6),  Fig.  368.  Considering  the  plank  to  be  a 
fixed  beam,  as  before,  and  assuming  the  load  on  the  wheel  to  be  2,000*, 
we  have 

(2,000  x  27) -r  8  =  6,750  inch  Ibs. 

for  the  maximum  bending  moment  at  A  and  also  at  the  center  of  the 
span,  it  being  a  negative  moment  at  A  and  a  positive  moment  at  the  cen- 
ter of  the  span. 

As  seen  from  above,  the  moment  -  25,920"*  due  to  the  traction 
engine  is  the  greatest  bending  moment  on  the  plank.  Assuming  each 
plank  to  be  3"  x  12",  we  have 


for  the  moment  of  inertia.     Then  for  the  maximum  fiber  stress  we  have 
/=  ^25,920  xli^-5-  27  =  1,440  Ibs. 

(see  Art.  53).  This  stress  is  for  live  load  alone.  The  dead  load — the 
weight  of  the  plank — will  raise  this  slightly.  The  1,440*  stress  given 
above  is  about  the  allowable  intensity,  but  owing  to  the  fact  that  lumber 
under-runs  in  thickness  and  that  the  thickness  is  reduced  also  from 
wear,  we  will  use  4"  x  12"  plank.  The  weight  of  the  plank  can  be  taken 
as  4.25*  per  ft.  B.  M.  Then  for  the  weight  per  ft.  of  a  4"  x  12"  plank 
we  have  4.25  x  4  =  17*.  Considering  a  fixed  beam,  as  before,  we  obtain 

(17x2J  )  + 12  =  7.17  foot  Ibs.  or  86  inch  Ibs. 

for  the  dead-load  bending  moment  at  the  supports  (see  Art.  69). 

Then  adding  this  to  the  above  maximum  live-load  moment  we  ob- 
tain 

25,920  +  86  =  26,006  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  floor  plank.  Now,  apply- 
ing Formula  D  of  Art.  53,  we  have 

/'  =  (26,006x2)  +64  =  812  Ibs. 

for  the  maximum  bending  stress  on  the  4"xl2"  floor  plank.  This  is 
rather  a  low  stress,  but  when  we  consider  the  wear  from  traffic  and  that 
some  plank  may  be  only  8"  wide  and  the  thickness  in  some  cases  may 
really  be  only  3^"  it  is  evident  that  4"  is  none  too  thick  and  hence  we 
will  use  that  thickness. 

It  is  true  that,  by  planking  for  heavy  loads,  planks  3"  thick  could 
be  used.  However,  there  is  no  question  but  that  it  would  have  to  be 
replaced  sooner  than  if  4"  plank  were  used  and  consequently  the  saving 
is  not  as  great  as  it  appears. 

225.  Design  of  Beams. — The  beams  will  be  about  2  7"  =  2^  apart. 
The  4"  floor  will  weigh  4x4^  =  17*  per  sq.  ft.  Then  we  have 
17x2^  =  38.25*  for  the  dead  load  on  a  beam  from  the  floor  planks  and 


542 


STRUCTURAL  ENGINEERING 


assuming  the  beam  to  weigh  25*  per  ft.  we  have  38.25  +  25  =  63.25,  say 
64#  for  the  total  dead  load  per  ft.  of  beam.     Then  we  have 

M=i-x.64xl82  x  12  =  31,104,  say  31,000  inch  Ibs. 
8 

for  the  maximum  dead-load  bending  moment. 

Prof.  F.  O.  Dufour  and  the  author,  working  independently  of  each 

other,  found  that  the  wheel  concentra- 
tion on  joists  (spaced  about  2' -3" 
to  2'  -  6"  apart)  supporting  wood 

floors   was  about  43   per  cent   of  the 

j  I       wheel  load.     To  be  on  the  safe  side 

i  we    will    assume    50    per    cent.      The 

~^  I       maximum  live  load  moment  will  occur 

Fig.  369  when  one  large  wheel  is  in  the  middle 

of   the    span    as    shown   in   Fig.    369. 
Then  we  have 


M"  =  ^--  x  9  x  12  =  270,000  inch  Ibs. 

18 

for  the  maximum  bending  moment.  Now,  adding  the  maximum  dead 
and  live  load  moments  together  we  have 

M'"  =  31,000  +  270,000  =  301,000  inch  Ibs. 

for  the  total  maximum  bending  moment  on  a  beam.  Then  we  obtain 
301,000*16,000  =  18.8  for  the  section  modulus.  This  calls  for  a  9"x21# 
I-beam.  The  design  of  the  bridge  is  shown  in  Fig.  370. 

Complete  Design  of  an  18-Ft.  Beam  Span  with  Concrete  Floor 

226.  Data.— 

Length  of  span  =  18'  —  0"  c.  to  c.  of  end  bearings. 
Roadway  =  16'-0". 

Specifications,  as  given  in  Art.  222. 

227.  Designing  of  the  Concrete  Floor  Slab. — The  I-beams  sup- 
porting concrete   floor   slabs   are   usually   spaced    from   2'— 3"  to   2'-6" 
apart  and  it  is  usual  to  limit  the  minimum  thickness  of  the  slab  to  about 
6"  and  the  fill  to  3"  of  gravel.     So  in  this  case  we  will  assume  the  slab 
to  be  6"  thick,  the  fill  3",  and  space  the  beams  as  shown  at  (a),  Fig. 
371.     The  weight  of  the  slab  per  sq.  ft.  of  floor  is  150x£  =  75#,  and 
the  weight  of  fill  per  sq.  ft.  is  120  x  J  =  30#,  making  in  all  75  +  30  =  105# 
per  sq.  ft.  of  floor.     Now  considering  a  transverse  strip  1  ft.  wide  as  a 
continuous  beam  over  eight  supports  we  have 

M  =  -J^xl05x2^52x  12  =  -673.7  inch  Ibs. 

for  maximum  bending  moment  on  slab  due  to  dead  load. 

As  is  seen,  the  large  wheel  of  a  traction  engine  will  practically 
extend  from  beam  to  beam  and  hence  the  maximum  live  load  on  the 
transverse  strip  considered  above  will  really  be  uniformly  distributed. 
We  know  that  this  load  will  be  distributed  along  the  joists  to  some 
extent.  Let  us  assume  that  it  is  distributed  for  2  ft.  along  the  joists. 
Then  we  have  the  live  load  as  shown  at  (6),  Fig.  371.  Now  consider- 


X 


ll 

fin 


543 


544 


STBUCTUEAL  ENGINEEEING 


ing  the  transverse  strip  to  be  a  fixed  beam  (which  is  a  safe  assumption) 
we  obtain 

M'  =  --^x5,000-x  2.25x12  =  11,250  inch  Ibs. 

for  the  maximum  bending  moment  on  the  slab  due  to  live  load.  Then 
adding  the  dead-  and  live-load  moments  together  we  have  11,923"*  for 
the  maximum  moment  on  the  slab.  This  moment  requires  that  the  con- 
crete slab  (6"  thick)  be  reinforced  with  |"  square  rods  spaced  about  18" 


(a) 


sooo 


Vigaft^rg^ifa*. 


(b) 


Fig.   371 


centers  top  and  bottom.  But  this  spacing  would  not  be  satisfactory  as 
the  rods  would  be  too  far  apart  to  properly  distribute  the  stress  through 
the  slab  and  consequently  we  will  reinforce  the  slab  with  \"  square  rods 
spaced  12"  apart  top  and  bottom  in  the  transverse  direction  and  two  J" 
square  rods  at  bottom  of  slab  between  beams  in  the  longitudinal  direc- 
tion. The  maximum  shear  on  the  slab  will  be  about  2,500  +  110  =  2,610*. 
The  cross-section  of  the  1-ft.  transverse  strip  including  the  reinforcing 
is  about  79n".  So  the  maximum  unit  shearing  stress  on  the  slab  is 
2,610  -5-  79  =  33*  (about),  which  is  quite  satisfactory. 

It  will  be  found  by  calculation  that  the  above  slab  is  safe  in  case 
the  beams  be  spaced  as  far  apart  as  3'  -  3". 

228.  Design  of  Beams.  —  Assuming  a  beam  to  weigh  25*  per  ft., 
the  total  dead  weight  on  each  intermediate  beam  will  be  (75x2.25)  + 
(30x2.25)  +25  =  261.25,  say  262*  per  ft.  of  beam.  Then  we  have 

M  =  |-  x  262  x  182  x  12  =  127,333,  say  127,000  inch  Ibs. 

o 

for  the  maximum  dead-load  bending  moment. 

Now,  assuming  that  each  intermediate  beam  supports  one-third  of 
a  wheel  instead  of  one-half  as  in  the  case  of  wood  floors  (see  Fig.  369), 
we  obtain 

x  9  x  12  =  180,000  inch  Ibs. 


for  the  maximum  moment  on  each  intermediate  beam  due  to  live  load. 
Adding  the  dead-  and  live-load  moments  together  we  have 
127,000  +  180,000  =  307,000  inch  Ibs. 


I 


i   I 
I  <* 


j 


545 


546  STEUCTUBAL  ENGINEEEING 

for  the  total  maximum  bending  moment.  Then  we  have  307,000  +  16,000 
=  19.2  for  the  section  modulus.  This  calls  for  a  9"x21#  I-beam,  which 
is  the  same  as  found  above  for  the  span  with  wood  floor.  The  outside 
beams  will  very  likely  be  subjected  to  about  two-thirds  of  the  live  load 
and  one-half  of  the  dead  load  carried  by  the  intermediate  beams.  This 
requires  that  the  outside  beams  be  9"xl5#  channels.  The  design  of 
the  bridge  would  be  the  same  as  shown  in  Fig.  370,  except  the  floor 
would  be  of  concrete.  Fig.  372  shows  the  design  of  beam  bridges  sup- 
porting concrete  floors  as  used  by  the  Iowa  Highway  Commission.  It 
will  be  seen  that  the  spans  given  in  Fig.  372  are  the  clear  span  in  each 
case. 

PONY  TRUSS  BRIDGES 

Complete  Design  of  a  50-Ft.  Pony  Truss  Bridge  with  Concrete  Floor 

and  without  Joists 

229.  Data.— 

Length     =6  panels  @  8' -4"  =  50'-  0"  c.c.  of  end  bearings. 

Height     =  6'  -  6"  c.c.  chords. 

Width       =  19'  -  3"  c.c.  trusses. 

Roadway  =  18'  -0". 

Specifications,  as  given  in  Art.  222. 

230.  Design  of  Concrete  Slab. — As  the  concrete  slab  is  continuous 
from  one  end  of  the  bridge  to  the  other  and  is  supported  directly  upon 
the  floor  beams    (no  joists)    it  is  really  a  continuous  beam  over   seven 
supports  and  there  is  no  objection  to  considering  the  slab  as  an  out  and 
out  continuous  beam  in  determining  the  maximum  moments  and  shears; 
but  sufficiently  accurate  results  will  be  obtained  by  considering  the  slab 
in  each  end  panel  as  a  beam  fixed  at  one  end  and  supported  at  the  other, 
and  the   slab   in  each  of  the  intermediate   panels   as   fixed   beams.      In 
determining  the  moments  and  shears  on  the  slab  under  either  assumption, 
a  longitudinal  strip  1   ft.  wide  is  considered.     In  the  case  of  live  load 
only  one-fourth  of  a  wheel  is  considered  to  be  carried  by  this  strip.     It 
is  found  from  calculation  that  a  slab  7J"  thick  reinforced  top  and  bot- 
tom longitudinally  with  f"  square  rods  spaced   9"  centers  is  sufficient 
for  all  panels  from  7'  -  0"  to  8'  -  6"  in  length  and  a  slab  8"  thick  with 
the  same  reinforcing  is  sufficient  for  all  panels  from  8'  — 6"  to  9"-0'  in 
length. 

231.  Design  of  Intermediate  Floor  Beams. — Taking  the  slab  as 
7J"  thick  and  the  fill  as  3"  the  dead  load  on  the  floor  beam  from  the 
slab  is   124x8.33  =  1,033*  per   ft.   of  beam  and   assuming  the  beam  to 
weigh  55#,  we  have  1,033  +  55  =  1,088*  for  the  total  dead  load  on  the 
floor  beam  per  ft.  of  length.     Then  we  have 

M  =^  x  1,088  x  19.252  x  12  =  604,753,  say  605,000  inch  Ibs. 
o 

for  the  maximum  bending  moment  on  the  floor  beam  due  to  dead  load. 
Placing  the  wheel  loads  as  shown  in  Fig.  373,  we  obtain 

M'  =  8,437x8.12x12  =  822,000  inch  Ibs. 
for  the  maximum  bending  moment  on  the  floor  beam  due  to  live  load. 


DESIGN  OF  SIMPLE  HIGHWAY  BEIDGES 


547 


Now  adding  the  above  moments  together  we  have 

•605,000  +  822,000  =  1,427,000  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  floor  beam.  Then  we 
have  1,427,000 •fl6,00Q  =  89.2  for  the  section  modulus.  This  calls  for  an 
I-18"x55*. 


3'    \I.5I.5' 

T 

n 

I 

;               j 

T 

1 

R=S437. 

3.G2' 

3.62' 

/9-3" 

Fig.  373 

232.  Design  of  End  Floor  Beams.  —  The  maximum  bending  mo- 
ment due  to  dead  load  is  about  one-half  of  the  dead-load  moment  on  the 
intermediate  beam   and  the  live-load  moment  is  exactly  the  same.     So 
for  the  total  maximum  bending  moment  on  an  end  floor  beam  we  have 

318,000  +  822,000  =  1,140,000  inch  Ibs. 

Then    we   have    1,140,000  +  16,000  =  71.2    for    the    section    modulus. 
This  also  calls  for  an  I  -  18"  x  55*. 

233.  Dead-Load  Stresses   in  Trusses.  —  From   Formula    (1)    of 
Art.  221  we  obtain 


Ibs. 

for  the  weight  of  metal  per  ft.  of  span  and  for  the  dead  load  from  the 
floor  we  have  124  x  18  =  2,232*  per  ft.  of  span.  Then  for  the  total  dead 
load  we  have  420  +  2,232  =  2,652*  per  ft.  of  span  or  1,326*  per  ft.  of 
truss. 

Then  for  a  panel  load  we  have  1,326x8.33  =  11,045,  say  11,- 
000*.  Practically  all  of  this  is  at  the  bottom  panel  points  and  hence 
we  will  so  consider  it. 


Ul    -1-57000       U 2+ 5  7000     U3 


LO1 


LI 


L2       (-4/ 
€>  PQnels 


L3 


=  5Oo 


LO 


Fig.  374 


Referring  to  Fig.  374  it  is  seen  that  the  stress  in  the  hangers  Ul-Ll 
and  U3-L3  is  a  panel  load,  or  11,000*.  The  stress  in  U3-L2  is  equal 
to  one-half  a  panel  load  multiplied  by  sec0  or  (11,000  +  2)  1.63  =  8,965, 
say  9,000*  compression. 

The  stress  in  diagonal  U1-L2  is  equal  to  one  and  one-half  of  a  panel 


548 


STRUCTURAL  ENGINEERING 


load  multiplied  by  sec0  or  (11,000)  1£  X  )  .63  =  26,895,  say  27,000*  ten- 
sion. 

The  stress  in  the  end  post  LO-Ul  is  equal  to  two  and  one-half  of  a 
panel  load  multiplied  by  sec0  or  (11,000)  2^x1.63  =  44,825,  say  45,000* 
compression.  The  post  U2-L2  does  nothing  but  stiffen  the  top  chord  and 
hence  the  only  load  coming  on  it  is  the  weight  of  the  metal  at  joint  £72, 
which  is  about  600*.  The  stress  in  bottom  chord  LO-L2  is  equal  to 
Rtan6=  (11,000)  2^x1.28  =  35,200,  say  35,000*  tension  (see  Art.  173). 

The  stress  in  top  chord  U1-U3  is  equal  to  (11,000)  4  x  1.28  =  56,320, 
say  57,000*  compression.  The  stress  in  bottom  chord  L2-L3  is  equal  to 
(11,000)  4|x  1.28  =  63,360,  say  64,000*  tension.  This  completes  the  de- 
termination of  the  dead-load  stresses  in  the  trusses  and  we  will  next 
take  up  the  determination  of  the  live-load  stresses. 

234.  Determination  of  Live-Load  Stresses  in  Trusses. — As  per 
specifications  (see  3  of  Art.  222)  the  uniform  live  load  is  100*  per  sq. 
ft.  of  roadway.  Then  for  the  panel  load  we  have 

W=  100x9x8.33  =  7,497  Ibs. 
We  have  the  following  for  determining  the  stresses  in  the  trusses: 

Tan0          =  1.28, 
Sec0  =  1.63, 

|  Wsec6   =2,036  Ibs., 
WtanO      =9,596  Ibs. 

Loading  panel  points  L5  and  L4  (Fig.  375)  we  have 
2,036x3  =  6,108,  say  6,000  Ibs. 


Ul     +38000* U2  +38000* U3 


LO 


U4 


US 


L3 
3 


8-4 


L4 
Z 

CD 


L5 
I 


SO'-O 


Fig.  375 

for  the  tensile  stress  in  diagonal  U3-L4.     Loading  panel  points  1/5,  L4 
and  L3  we  have 

2,036x6  =  12,216,  say  12,000  Ibs. 

for  the  maximum  compression  in  diagonal  U3-L2.     Loading  panel  points 
L5  to  L2  we  have 

2,036x10  =  20,360,  say  20,000  Ibs. 

for  the  maximum  tensile  stress  in  diagonal  U1-L2.    Loading  panel  points 
L5  to  LI  we  have 

2,036x15  =  30,540,  say  30,000  Ibs. 

for  the  maximum  compression  in  end  post  Ul-LO. 

As  is  readily  seen,  the  stress  in  hangers  C71-L1,  C73-L3,  and  U5-L5 
is  tension  and  equal  to  a  panel  load  of  7,497  Ibs.     There  is  no  live-load 


DESIGN  OF  SIMPLE  HIGHWAY  BKIDGES 


549 


stress  in  the  posts   U2-L2  and  Z74-L4.     Loading  all  points,  L5  to  LI, 
we  have 

9,596x2^  =  23,990,  say  24,000  Ibs. 
a 

for  the  stress  in  bottom  chord  LO-L2  and  9,596  x  4^  =  43,182,  say  43,000* 
for  the  stress  in  bottom  chord  L2-L4.     In  a  similar  manner  we  obtain 

9,596x4  =  38,383,  say  38,000  Ibs. 

for  the  stress  in  top  chord  U1-U3.  These  chord  stresses  can  be  deter- 
mined more  readily  by 
direct  proportion.  For 
example,  the  dead-load 
panel  load  is  11,000* 
and  the  live-load  panel 
load  is  7,49  7*,  and 
hence  multiply- 
ing (7,497 -=-11,000) 
by  the  dead-load  stress 
in  any  chord  member 

we    would    obtain    the  R=  14-800' 

live-load    stress    in   it. 

Thus,  for  the  live-load  Fig.  376 

stress  in  L2-L3  we  have 

/  7,49  7 


1                              13.25' 

74.25'                        3' 

2' 

r 

* 

^ 

N. 

~ 

3\ 

—     T~ 

§ 
S 

J 

11,000 


x  64,000  =  43,650,  say  43,000  Ibs. 


We  now  have  all  of  the  stress  due  to  the  uniform  live  load  deter- 
mined and  next  we  will  consider  the  stress  caused  by  the  engine.  Placing 
the  large  wheels  transversely  over  a  floor  beam  as  shown  in  Fig.  376  and 
taking  moments  about  A  we  obtain 

(20,000  x  14.25) -r  19.25  =  14,800,  say  15,000  Ibs. 

for  the  reaction  at  J5,  which  is  the  maximum  stress  in  hangers  C71-L1, 
[73-L3,  and  Z75-L5.  This  is  greater  than  the  stress  caused  by  the  uni- 
form load  and  consequently  it  will  be  considered  in  designing  these 
members. 

By  placing  the  wheels  as  shown  in  Fig.  377  we  obtain  a  stress  of 
[  (22,500  x  21.34) -r  50]  1.63  =  15,600,  say  16,000*  compression  in  diag- 
onal t/3-L2,  which  is  greater  than  the  stress  caused  by  the  uniform  load. 
In  a  similar  manner,  by  placing  a  large  axle  at  L4  and  the  smaller  one  to 
the  right,  we  obtain  a  tensile  stress  of  [  (22,500  x  13) -i- 50]  1.63  =  9,500, 
say  9,000*  in  diagonal  C73-L4.  This  stress  is  also  larger  than  that  ob- 
tained by  the  use  of  the  uniform  live  load  and  hence  will  be  considered 
in  designing  the  diagonals  U3-L2  and  £75-L4. 

The  hangers  £71-L1,  etc.,  and  the  diagonals  U3-L2  and  173-L4  are 
the  only  members  that  are  subjected  (as  ascertained  by  trial)  to  greater 
stress  from  the  engine  than  from  the  uniform  live  load  and  consequently 
the  stress  in  the  other  members  due  to  the  engine  will  not  be  considered. 
In  the  case  of  shorter  spans  the  stresses  due  to  the  engine  will  influence 
the  design  of  more  members.  In  fact  it  is  always  advisable  to  determine 


550 


STRUCTURAL  ENGINEERING 


the  stresses  due  to  the  engine  in  most  all  the  members  of  a  truss  to  make 
sure  that  the  absolute  maximum  stress  is  obtained  in  each  case.  Influence 
lines  can  be  used  to  an  advantage  in  determining  the  stress  in  the  truss 
and  especially  in  the  case  of  the  engine  load. 

The  maximum  dead-load  stresses  in  the  truss  are  given  in  Fig.  374 

3.66* 


LO 


U3 


LI 


L6 


/4--O 


-  O 


Fig.  377 

and  the  maximum  live-load  stresses  in  Fig.  375.  Combining  these  we 
obtain  the  stress  diagram  shown  at  the  top  of  Fig.  378. 

235.  Designing  of  the  Sections  of  Truss  Members. — The  work 
of  designing  the  truss  members  is  practically  the  same  as  previously 
given  for  railroad  bridges.  The  vertical  posts  and  hangers  as  far  as 
stresses  are  concerned  could  be  very  light  section,  but  4 — Ls  3"  x  2J"  x  J" 
and  a  plate  7"  x  \"  is  about  as  light  a  section  as  can  be  used  for  these 
members  and  obtain  good  details. 

Diagonals  U3-L2  must  be  designed  to  carry  25,000*  compression  or 
12,500*  tension.  The  effective  length  of  this  member  is  about  126". 
Then  using  2— Ls  3J"x2J"  xi"  =  3.76n"  (3J"  legs  vertically),  we  have 


L   126 


=  113 


r  1.12 

and  hence  we  obtain  16,000-70x113  =  8,090*  for  the  allowable  unit 
compression  stress.  Then  25,000 -=-  8,090  =  3.09n"  is  the  required  area 
for  compression  and  12,500  -r  16,000  =  0.78n"  is  area  required  for  ten- 
sion. If  less  than  3^"x  2^"  angles  be  used  L/r  would  be  too  great,  so  we 
will  use  2— Ls  3£"  x  2J"  x  J"  for  diagonal  U3-L2.  The  designing  of  the 
other  diagonals  and  the  bottom  chord  is  simply  a  matter  of  determining 
the  correct  net  section  in  each  case  as  previously  explained  in  the  design 
of  railroad  bridges. 

The  maximum  stress  in  the  top  chord  is  95,000*.  Dividing  this  by, 
say  12,000  Ibs.,  we  obtain  about  Sn"  for  the  approximate  area  of  cross- 
section  of  the  chord.  Then  assume  the  top  chord  to  be  composed  of  the 
following : 

2—[s  6"x8*     =4.76°" 

1— Cov.  i: 


7.76n" 

The  radius  of  gyration  of  this  section  with  reference  to  the  hori- 
zontal axis  is  2.38"  and  with  reference  to  the  vertical  axis  it  is  4.08". 


551 


552 


553 


554  STKUCTUEAL  ENGINEEEING 

Considering  the  chord  to  fail  vertically  the  length  would  be  taken  as 
one  panel  length,  or  100".     Then  we  have 


16,000-  70  =  13,060  Ibs. 

for  the  allowable  unit  stress.     Dividing  this  into  the  stress  we  have 
95,0004-13,060  =  7.27  sq.  ins. 

If  the  chord  be  considered  with  reference  to  the  vertical  axis  the  length 
should  be  taken  as  two  panel  lengths,  or  200".     Then  we  obtain 


16,000-70  =12,570  Ibs. 

\4.0S/ 

for  the  allowable  unit  stress.      Dividing  this  into  the  stress  we  obtain 
95,000  +  12,570  =  7.56  sq.  ins. 

This  shows  that  the  assumed  chord  section  is  about  correct  and  hence 
will  be  used.  The  end  posts  are  designed  in  a  similar  manner  and  the 
laterals  are  designed  to  transmit  a  wind  load  of  300*  per  ft.  of  span. 
Complete  details  of  the  span  are  shown  in  Fig.  378. 

236.  Long  Pony  Truss  Spans.  —  These  spans  are  designed  in  the 
same  manner  as  shown  above  for  the  50-ft.  span.     The  ends  of  all  spans 
65  ft.  and  over  in  length  should  be  pin-bearing  and  have  a  cast  rocker  at 
one  end  and  a  cast  fixed  shoe  at  the  other  and  have  brackets  stiffening 
the  top  chord,  as  shown  in  Figs.  379  and  380. 

THROUGH  TRUSS  BRIDGES 
Complete  Design  of  a  112-Ft.  Through  Pratt  Truss  Span 

237.  Data.— 

Length     =    7  panels  @  16'-0"  =  112'-0"  c.c.  end  pins. 

Height     =  20'  -  0"  c.c.  chords. 

Width      =17'-  0"  c.c.  trusses. 

Roadway=16'-0". 

Specifications,  as  per  Art.  222. 

238.  Designing  of  the  Joists  and  Concrete  Slab  is  the  same  as 
previously  shown  for  beam  bridges.     The  joists  will  be  9"x21*  Is  and 
9"  x  13.25*  [s.    The  slab  will  be  6"  thick  reinforced  with  \"  square  rods 
spaced  12"  centers.    The  fill  will  be  3"  thick. 

239.  Designing  of  the  Intermediate  Floor  Beams.  —  The  weight 
of  the  slab  and  fill  is  105  Ibs.  per  sq.  ft.  of  roadway.     The  joists  can  be 
taken  as  weighing  21+2^  =  8.4  Ibs.  per  sq.  ft.  of  roadway.     Then  from 
the  slab,  fill,  and  joists  we  have  (105  +  8.4)  16  =  1,814*  for  the  load  per 
ft.  of  floor  beam  and  assuming  the  floor  beam  to  weigh  65*  per  ft.  we  have 
1,814  +  65  =  1,879*  for  the  total  dead  load  on  the  floor  beam  per  ft.  of 
length.    Then  for  the  maximum  bending  moment  on  the  floor  beam  due  to 
dead  load  we  have 

M  =  i  x  1,879  x  rf  x  12  =  814,500  inch  Ibs. 
o 


DESIGN  OF  SIMPLE  HIGHWAY  BEIDGES 


555 


Placing  the  wheel  loads  as  indicated  in  Fig.  381  and  assuming  that 
-f$  of  the  smaller  wheels  is  transmitted  to  the  floor  beam  under  the  larger 
wheels,  we  have 

M'  =  9,520  x  7  x  12  =  800,000  inch  Ibs. 
for  the  maximum  live-load  bending  moment  on  the  beam. 

4*0"     3L0' 


5-0' 


//-O 


ieLo 


Fig.  381 

Adding  the  above  dead-  and  live-load  moments  together  we  have 

814,500  +  800,000  =  1,614,500  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  beam.  Then  we  have 
1,614,500  +  16,000  =  101  for  the  section  modulus.  This  calls  for  a 
20"x65*  I. 

The  designing  of  the  end  floor  beams  is  practically  the  same  as  for 
the  intermediate  beams.  The  live  load  is  exactly  the  same  and  the  dead 
load  is  about  one-half  as  much  as  for  the  intermediate  beams. 

240.    Determination  of  the  Dead-Load  Stresses  in  Trusses. — 

From  (2)  of  Art.  221  we  obtain  2.8x112  +  280  =  594*  for  the  weight  of 
metal  per  ft.  of  span  and  for  the  weight  of  the  concrete  slab  and  the 
fill  we  have  (75  +  30)  16  =  1,680*  per  ft.  of  span,  making  in  all  594  + 
1,680  =  2,274*  per  ft.  of  span  or  1,137,  say  1,130  Ibs.  per  ft.  of  truss. 
Then  we  have 

W=  16x1,130  =  18,080  Ibs. 

for  the  panel  load.  The  expressions  for  the  dead-load  stresses  are  given 
on  the  right  half  of  the  truss,  Fig.  382,  and  the  actual  stresses  are  ffiven 


Fig.  382 

on  the  left  half.  About  one-third  of  the  weight  of  metal  is  assumed  to  be 
at  the  top  chord  joints.  This  is  about  one-twelfth  of  the  total  dead  load 
and  hence  the  distribution  shown  is  assumed.  For  determining  the  stresses 
we  have  the  following  values : 


STRUCTURAL  ENGINEERING 
Tan0=0.8. 


556 


WtanO=  14,464  Ibs. 
Wsec0=  23,142  Ibs. 

The  stresses  are  determined  as  explained  in  Art.  173. 

241.  Determination  of  Live-Load  Stresses  in  Trusses. — We  will 
first  consider  the  80-lb.  load  specified  in  (3)  of  Art.  222.  For  the  panel 
load  we  have 

P  =  80x8xl6  =  10,2401bs. 

For  determining  the  stresses  we  have  the  following  values: 
P  tan0  =  10,240  x  0.8  =  8,192  Ibs. 

ip  =  1,463  Ibs. 
i-Psec0  =  1,872  Ibs. 
Loading  panel  points  G  and  F  (Fig.  383)  we  obtain 


Fig.  383 

1,872  x  3  =  5,616,  say  5,600  Ibs. 

for  the  maximum  compressive  stress  in  diagonal  fE.    Loading  panel  points 
G,  F,  and  E  we  obtain 

1,872  x  6  =  11,232,  say  11,000  Ibs. 

for  the  maximum  tensile  stress  in  diagonal  dE.     Loading  panel  points 
G  to  D  we  obtain 

1,872  x  10  =  18,720,  say  18,800  Ibs. 

for  the  maximum  tensile  stress  in  diagonal  cD.     Loading  panel  points 
G  to  C  we  obtain 

1,872  x  15  =  28,080,  say  28,000  Ibs. 

for  the  maximum  tensile  stress  in  diagonal  bC.     Loading  panel  points 
G  to  B  we  obtain 

1,872  x  21  =  39,312,  say  39,000  Ibs. 

for   the  maximum   compressive   stress  in   end   post   bA.      Loading  panel 
points  G  to  E  we  obtain 

1,463  x  6  =  8,778,  say  8,800  Ibs. 


DESIGN  OF  SIMPLE  HIGHWAY  BRIDGES 


557 


for  the  maximum  compressive  stress  in  post  dD  and  loading  panel  points 
G  to  D  we  obtain 

1,463  x  10  =  14,630,  say  14,000  Ibs. 

for  the  maximum  compressive  stress  in  post  cC.  The  stress  in  hanger  bB, 
as  is  evident,  is  equal  to  a  panel  load  which  is  given  above  as  10,240*. 
We  now  have  the  stresses  in  the  web  members  determined  and  we  will 
next  determine  the  stresses  in  the  chords.  The  maximum  stress  in  the 
chords  occurs  when  the  span  is  fully  loaded.  For  maximum  stress  in 
bottom  chord  A-C  we  have 

8,192  x  3  =  24,576,  say  24,500  Ibs. 
For  maximum  stress  in  top  chord  be  and  bottom  CD  we  have 

8,192  x  5  =  40,960,  say  41,000  Ibs. 
For  the  maximum  stress  in  top  chord  c-f  and  bottom  chord  DE  we  have 

8,192  x  6  =  49,152,  say  49,000  Ibs. 

We  now  have  the  stresses  due  to  the  uniform  load  determined  and  we 
will  now  consider  the  stresses  due  to  the  engine.  Placing  the  larger  wheels 
over  a  floor  beam  as  shown  in  Fig.  384 

we     obtain     (using    the     concentration 

shown  in  Fig.  381) 


1 9 

(11,560  x  $)||  =  16,320,  say  16,300  Ibs. 

for  the  maximum  end  reaction  on  a  floor 
beam  which  is  also  the  maximum  live- 
load  stress  in  each  hanger  and  hence 
this  will  be  used  in  designing  the  hangers. 

Placing  the  wheel  loads  longitudinally,  as  shown  in  Fig.  385,  and 
transversely,  as  shown  in  Fig.  384,  we  obtain  the  concentrations  indicated 


Fig.   384 


Fig.   385 


at  E  and  F.     For  the  shear  in  panel  DE  due  to  these  concentrations  we 
have 


This  is  equal  to  the  maximum  stress  in  post  dD  due  to  the  engine,  and 
multiplying  the  same  by  sec#  we  obtain 

8,370  x  1.28  =  10,714,  say  10,700  Ibs. 
for  the  maximum  stress  in  diagonal  dE  due  to  the  engine.     But  as  these 


558 


STRUCTURAL  ENGINEERING 


stresses  (in  dD  and  dE)  due  to  the  engine  are  less  than  produced  by  the 
80-lb.  uniform  load  we  need  not  consider  further  the  stresses  produced 
by  the  engine. 

As  is  seen,  the  hangers  are  the  only  truss  members  receiving  maxi- 
mum stress  from  the  engine  load. 

242.  Determination  of  Wind  Stresses  in  the  Laterals.  —  Let  us 
first  consider  the  bottom  laterals.  The  load  as  given  in  4  of  Art.  222 
is  300*  per  ft.  of  span.  Then  for  the  panel  load  we  have 

P  =  300  x  16  =  4,800  Ibs. 
For  determining  the  stresses  we  have  the  following  values  : 


=  4,800,seco>  =  1. 


and      Pseco>  = 


H 


Fig.  386 

Loading  panels  from  h  to  e  (Fig.  386)  we  obtain 
939  x  6  =  5,634,  say  5,600  Ibs. 

for  the  maximum  stress  in  lateral  dE.  Loading  panel  points  h  to  d  we 
obtain 

939  x  10  =  9,390,  say  9,400  Ibs. 

for  the  stress  in  lateral  cD.     Loading  panel  points  h  to  c  we  obtain 

939  x  15  =  14,085,  say  14,000  Ibs. 
for  the  stress  in  lateral  bC.    Loading  panel  points  h  to  b  we  obtain 

939x21  =  19,719,  say  20,000  Ibs. 

for  the  stress  in  diagonal  aB.  We  now  have  all  the  stresses  necessary 
for  designing  the  bottom  laterals  determined  (see  Art.  177)  and  we  will 
next  consider  the  top  laterals.  The  load  as  given  in  4  of  Art.  222  is  100* 
per  ft.  of  span.  Then  for  the  panel  load  we  have 

P'  =  100  x  16  =  1,600  Ibs.  and  }  P'  secco  =  313. 
Loading  panel  points  g  to  e  (Fig.  387)  we  obtain 

6  x  313  =  1,878,  say  1,900  Ibs. 

for  the  stress  in  top  lateral  dE.     Loading  panel  points  g  to  d  we  obtain 
10  x  313  =  3,130,  say  3,100  Ibs. 


DESIGN  OF  SIMPLE  HIGHWAY  BEIDGES 


559 


for  the  stress  in  top  lateral  cD.     Loading  panel  points  g  to  c  we  obtain 
15  x  313  =  4,695,  say  4,700  Ibs. 

for  the  stress  in  top  lateral  bC.     The  maximum  stress  will  occur  in  strut 
dD  when  panel  points  g  to  d  are  loaded.     The  half  of  the  panel  load 


D 
Fig.   387 


applied  at  D  does  not  affect  the  strut  and  hence  the  maximum  stress  in 
it  is  equal  to 

(f  P'  +  J  P')  =  f  x  1,600  +  4  x  1,600  =  2,172,  say  2,200  Ibs. 
Loading  panel  points  g  to  c  we  obtain 

^  P'  +  i  p'  \  =  3^)85,  say  3, 


000  Ibs. 


for  the  maximum  stress  in  strut  cC. 
243.    Stress   in  Portal.—  The 

maximum  stress  in  the  portal  mem- 
bers will  occur  when  all  of  the  panel 
points  are  loaded.  The  forces  on  the 
portal  will  be  as  shown  in  Fig.  388  : 

P'  =  1,600  Ibs. 

R  =  2P'  =  shear   in   panel    be 
when  all  panel  points  are  loaded. 


(£*%>)=  4000* 


c^-Qoo* 


»  »  Fig.  388 

Assuming  the  end  posts  fixed  and  taking  moments  about  0  we  obtain 

V=  (4,800  x  17.05)  ^  17  =  4,814,  say  4,800  Ibs. 
Now  assuming  0  =  45°  we  obtain 

4,800  x  1.41  =  6,768,  say  7,000  Ibs. 

for  the  stress  in  each  of  the  members  bD  and  bE,  the  same  being  tension 
in  one  and  compression  in  the  other. 

Taking  moments  about  D  we  obtain 

(2,400  x  8.55  +  4,000  x  8.5)  -r  8.5  =  6,414,  say  6,400  Ibs. 

for  the  compressive  stress  in  member  ab.     Taking  moments  about  E  we 
obtain 


d 


n*  «*% 

>|i  vf  J°!M|\« 


c  S*  3     ^5               -f- 

\  _ 

IQ 

is  - 

.  . 

T  F 

- 

Q 

*c 

V; 

i               ?r 

ji- 

__ 

* 

•- 

*  *in 

^ 

M                                                                                            ^j 

!       v^ 

isl 

KJ 

658^5 

it 


*?..  /'• ;  / 


/ 


-N 


X 


\ 


/ 


v§    . 
"^ 


7*f??*&/?j 


560 


DESIGN  OF  SIMPLE  HIGHWAY  BRIDGES  561 

(2,400  x  8.55  +  800  x  8.5)  -h  8.5  =  3,214,  say  3,200  Ibs. 

for  the  tensile  stress  in  member  be.  This  completes  the  calculations  of 
the  stresses  in  the  structure  and  the  stresses  can  be  written  on  the  stress 
sheet  Fig.  389. 

244.  Designing  of  Sections  of  the  Truss  Members. — Intermedi- 
ate Post.  The  length  of  each  of  these  posts  is  20  ft.  or  240  ins.  L/r 
should  not  exceed  125.  Then  we  have  240  -h  125  =  1.92  for  the  minimum 
allowable  value  of  r.  Using  this  value  of  r  we  obtain 

16,000  -  70  x  125  =  7,250  Ibs. 

for  the  allowable  unit  stress.  Dividing  this  into  the  stress  in  post  U2-L2 
(Fig.  389),  we  obtain 

34,000-^7,250  =  4.69  sq.  ins. 

for  the  required  area.  It  is  seen  from  this  that  two  6"  channels  will 
be  satisfactory  for  section.  The  5"  channels  can  not  be  used  as  the  r  is 
too  small  except  in  the  case  of  the  lightest  weight  which  have  not  enough 
section.  Considering  2— [s  6"x8#  =  4.76n",  we  obtain 


16,000-70x^  =  8,821  Ibs. 

for  the  allowable  unit  stress,  and  dividing  this  into  the  stress  in  post 
U2-L2,  we  obtain 

34,000  +  8,821  =  3.85  sq.  ins. 

for  the  required  area  of  cross-section,  which  is  0.91n/'  less  than  contained 
in  the  two  channels ;  but  these  channels  will  be  used  as  they  are  nearest 
to  the  theoretical  requirement,  and  for  the  same  reason  these  channels 
will  be  used  for  the  other  intermediate  posts. 

Top  Chord.  Assuming  12,000*  as  the  unit  stress  and  dividing  this 
into  the  maximum  stress  in  the  top  chord,  we  obtain  136,000^12,000  = 
11. 3n"  for  the  approximate  area  of  cross-section. 

Let  us  assume  the  following  section: 

1— cov.     12"xJ"  =    3.00n" 
2— [s  9"  x  13.25*  =   7.78n" 

10.78n" 
Taking  r  as  0.4  of  the  depth  of  the  chord  we  obtain 

p  =  16,000  -  70  ( ^- )  =  12,267 


for  the  allowable  unit  stress.  Dividing  this  into  the  maximum  stress  in 
the  top  chord  we  obtain 

136,000  *12,267  =  11.1  sq.  ins. 

for  the  required  area  of  cross-section,  which  is  only  0.32n"  more  than 
the  above  assumed  section.  The  actual  r  for  the  section  is  practically 
3.6,  the  same  as  assumed,  so  the  above  assumed  section  will  be  used  for 


562  STRUCTURAL  ENGINEERING 

the  top  chord  throughout.  The  area  required  for  U1-U2  is  less  than 
this  section,  but  as  minimum  thickness  is  used  the  area  can  not  be  changed 
without  changing  the  depth  of  the  chord  and  hence  the  same  section 
throughout  is  used. 

End  Post.     Let  us  assume  the  following  section  for  the  end  post: 

1— Cov.    12"xJ"  =    3.00n" 
2—  [s  9"x20#       =11.76*" 

14.76n" 

The  radius  of  gyration  about  the  vertical  axis  is  about  3.8  and 
about  the  horizontal  axis  it  is  about  3.5.  As  a  collision  strut  is  used  we 
need  consider  the  strength  only  with  regard  to  the  vertical  axis.  Sub- 
tracting the  depth  of  the  portal  we  have 

25.6-8.5  =  17.1  ft.,  or  205  ins. 

for  the  effective  length  of  the  end  post. 
Then  we  hr,ve 


p  =  16,000  -  70  (  y^  )  =  12,220  Ibs. 

for  the  allowable  unit  stress  for  direct  compression. 

Referring  to  Fig.  387,  it  is  seen  that  the  maximum  moment  on  the 
post  occurs  at  the  bottom  of  the  portal  and  is  equal  to 

2,400x8.55x12  =  246,240  inch  Ibs. 

The  moment  of  inertia  about  the  vertical  axis  is  about  220  and  the  dis- 
tance to  the  extreme  fiber  is  6  ins.     Using  these  values  we  obtain 

246,240 


220 


x  6  =  6,715  Ibs. 


for  the  unit  stress  due  to  the  cross  bending. 
For  the  direct  unit  stress  we  have 

108,000 -r  14.76  =  7,317  Ibs. 
Adding  this  to  the  bending  stress  we  have 

6,715 +  7,317  =  14,032  Ibs. 

for  the  total  combined  unit  stress.  The  allowable  unit  stress  given  above 
(12,220  Ibs.)  can,  according  to  the  specifications  (see  16  of  Art.  222), 
be  increased  25  per  cent  in  the  case  of  combined  stress.  So  for  the 
allowable  unit  stress  we  have 

12,220x1.25  =  15,275  Ibs. 

which  is  1,243*  greater  than  the  actual  combined  stress,  so  the  assumed 
section  is  a  little  larger  than  required,  but  it  will  be  used  as  the  section 
cannot  be  easily  reduced. 

The  designing  of  the  other  members  of  the  structure  is  mostly  a 
matter  of  determining  required  net  areas,  which  is  fully  explained  pre- 


563 


564 


STRUCTURAL  ENGINEERING 


viously  in  the  design  of  railroad  bridges.     After  the  sections  are  designed 
the  stress  sheet,  Fig.  389,  can  be  completed  as  shown. 

The  general  details  of  the  span  are  shown  in  Fig.  390.  These 
details  will  be  readily  understood  if  the  work  on  railroad  bridges,  pre- 
viously given,  is  thoroughly  studied. 

Complete    Design    of   a    162-Ft.    Through   Pin-Connected    Curved 
Chord  Pratt  Truss  Highway  Bridge 

245.  Data.— 

Length     =  9  panels  @  18'  -  0"  =  162'  -  0"  c.c.  end  pins. 

Height     =  20'  -  0"  at  hip  and  27'  -  10^"  at  center  (see  Fig.  392) 

Width       =  17'  —  6"  c.c.  of  trusses. 

Roadway  =  16'-0". 

Specifications,  as  per  Art.  222. 

246.  The  Designing  of  the  Floor  System  is  the  same  as  pre- 
viously shown   (see  Arts.   227,  228,  and  239). 

247.  Determination  of  Dead-Load  Stresses  in  Trusses. — From 
(2)  of  Art.  221  we  have 

2.8x162  +  280  =  734  Ibs. 

for  weight  of  metal  per  ft.  of  span  or  367  Ibs.  per  ft.  of  truss.     For  the 
weight  of  slab  we  have 

75x8  =  600  Ibs. 

per  ft.  of  truss,  and  for  the  weight  of  fill  we  have 

30x8  =  240  Ibs. 

per  ft.  of  truss.     Now,  considering  one-third  of  the  weight  of  the  metal 
at  the  top  chord  joints  and  two-thirds  at  the  bottom  chord  joints,  we  have 


f^pl  18  =  2,202,  say  2,200  Ibs. 

for  the  panel  load  on  each  top  chord  joint  and 

[§(367)  +  600  +  240]  18  =  19,524,  say  19,500  Ibs. 


for  the  panel  load  on  each  bottom  chord  joint.  Now  having  the  panel 
loads  computed  the  stresses  are  readily  determined  either  analytically  or 
graphically  in  the  same  manner  as  previously  shown  for  railroad  bridges 
(see  Arts.  189  and  195). 


DESIGN  OF  SIMPLE  HIGHWAY  BRIDGES  565 

248.  Determination  of  Live-Load  Stresses   in  Trusses. — The 
uniform  live  load  produces  maximum  stress  in  all  members  except  the 
hangers,  in  which  case  the  engine  produces  the  maximum  stress.     The 
uniform  live  load  as  per  3  of  Art.  222  is  70  Ibs.  per  sq.  ft.  of  roadway. 
Then  we  have 

P  =  70x8x18- 10,080  Ibs. 

for  the  panel  load. 

The  maximum  live-load  stresses  in  the  chords  occur  when  the  span 
is  fully  loaded  and  hence  are  directly  proportional  to  the  dead-load 
stresses  in  same  and  can  be  determined  very  quickly  by  multiplying  each 
dead-load  stress  by  the  live-load  panel  divided  by  the  dead-load  panel,  or 
the  live-load  stresses  in  the  chords  can  be  determined  as  explained  in 
Art.  189. 

The  maximum  stresses  in  the  web  members  due  to  the  uniform  load 
are  determined  in  very  much  the  same  manner  as  previously  shown  for 
railroad  bridges  except  that  in  the  latter  case  wheel  loads  were  used. 

The  maximum  tensile  stress  in  diagonal  eF  occurs  when  panel  points 
K,  H,  G,  and  F  are  loaded.  (See  Fig.  391.)  The  shear  in  panel  E-F 
iw  then  equal  to  ^-P  and  hence  the  stress  in  the  diagonal  is  equal  to 
1f-Psec63.  The  maximum  compressive  stress  in  post  eE  will  occur  at  the 
same  time.  This  stress  is  equal  to-1^0-  P-F2  where  F2  represents  the 
vertical  component  of  the  stress  in  top  chord  ed,  which  is  equal  to 
(l<l>-Px4d/h3)  tan</>2  (moments  about  E). 

The  maximum  tensile  stress  in  diagonal  dE  will  occur  when  panel 
points  K  to  E  are  loaded  and  is  equal  to  (-^P— F2)  sec  62.  The  maxi- 
mum compressive  stress  in  post  dl)  will  occur  at  the  same  time  and  is 
equal  to-^-P  — Fl,  where  V\  represents  the  vertical  component  of  the 
stress  in  top  chord  cd,  which  is  equal  to  (^5-Px3d/h2)  tan</>l. 

The  maximum  tensile  stress  in  diagonal  cD  occurs  when  panel  points 
K  to  D  are  loaded  and  is  equal  to  (-g^P—  VV)  sec.  01.  The  maximum 
compressive  stress  in  post  cC  occurs  at  the  same  time  and  is  equal  to 
-V  P  —  V,  where  V  represents  the  vertical  component  of  the  stress  in  top 
chord  cb,  which  is  equal  to  (\l-Px2d/hl)  tan  6.  The  maximum  tensile 
stress  in  diagonal  bC  occurs  when  panel  points  K  to  C  are  loaded  and  is 
equal  to  (-%8-P-F)  sec.  9. 

The  maximum  stress  in  hanger  bB,  which  is  due  to  the  engine  load, 
is  determined  as  previously  explained  (see  Art.  241). 

The  maximum  tensile  stress  in  counter  fG  will  occur  when  panel 
points  K,  H,  and  G  are  loaded,  and  is  equal  to  (JP-F3)  sec  03,  where 
F3  represents  the  vertical  component  of  the  stress  in  top  chord  fg. 

The  maximum  tension  in  the  posts  can  be  determined  as  previously 
explained  for  railroad  bridges  (see  Art.  190,  pages  422-426). 

249.  Stresses  in  Portals  and  Lateral  System. — From  (4)  of  Art. 
222  the  wind  load  on  the  bottom  laterals  is  300*  per  ft.  of  bridge  and 
100*  per  ft.  of  bridge  on  the  top  laterals.     Then  we  have  300x18  = 
5,400#  for  the  panel  load  on  the  bottom  laterals  and  100  x  18  =  1,800* 
for  the  panel  load  on  the  top  laterals.     The  stresses  are  determined  as 
explained  in  Art.  242  (also  see  Arts.  177-179). 


566  STRUCTURAL  ENGINEERING 

250.  Designing  of  the  Sections. — The  sections  of  the  truss  mem- 
bers  and   lateral   bracing   are   determined   as   previously   explained    (see 
Arts.  197,  198,  199,  200,  and  244). 

After  the  stresses  and  sections  are  determined  the  stress  sheet,  Fig. 
392,  can  be  made. 

251.  Details. — After   the   stress    sheet   is    completed,   the   general 
drawing,  Fig.   393,  can  be  made.      The  calculations  for  the  details  are 
practically  the  same  as  previously  shown  for  railroad  bridges   (see  Art. 
203).     The  student  should  verify  the  details  shown  in  Fig.  393. 

252.  Truss  Bridges  with  Wood  Floors  are  designed  in  the  same 
manner  as  bridges  with  concrete  floors.      The  total  dead  load  is  less  than 
for  bridges  with  concrete  floors,  but  the  live  load  is  the  same.     The  first 
cost  is  less,  but  in  most  cases  bridges  with  concrete  floors  will  prove  to 
be  the  cheaper  in  the  end. 

It  is  the  opinion  cf  the  author  that  all  highway  bridges  should  be 
designed  heavy  enough  to  carry  concrete  floors  even  if  wood  floors  are 
used,  as  the  county  authorities  usually  insist  on  putting  on  the  concrete 
sooner  or  later,  and  if  the  bridge  is  not  heavy  enough  to  carry  a  concrete 
floor  it  will  have  to  be  replaced  by  a  new  bridge,  which  practically  means 
the  loss  of  the  first  bridge. 


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andcashngrs. 

All 'rivets       '' 


JrANDARD /6?ft.  THRO.P/HCONft.  TfrUSS H/GHWAY BRIDGE. 
Concrete  floor -/6&'/ 


Fig.    393 


569 


CHAPTER  XIII 


SKEW  BRIDGES,  BRIDGES  ON  CURVES,  ECONOMIC  HEIGHT 
AND  LENGTH  OF  TRUSSES,  AND  STRESSES  IN  PORTALS 

253.  Skew  Bridges. — Bridges  crossing  over  streams,  railroads, 
streets,  and  roads  at  an  angle  are  usually  built  on  the  skew.  This  is 

done  in  the  case  of  crossings 
over  streams  in  order  to  ob- 
tain piers  parallel  to  the  flow 
of  the  stream  and  in  the 
other  cases  to  obtain  mini- 
mum span  lengths. 

Skew  deck  plate  girder 
railroad  bridges  are  usually 
constructed  as  shown  at  (a), 
Fig.  394,  through  girder 
spans  as  shown  at  (6),  truss 
spans  as  shown  at  (c) 
and  (d).  The  case  shown 
at  (c)  is  where  the  skew  is 
a  full  panel  length  and  the 
case  shown  at  (d)  is  where 
the  skew  is  less  than  a  panel 
length. 

Skew  highway  bridges 
are  constructed  very  similar 
to  skew  railroad  bridges,  as 
far  as  the  skew  is  concerned. 
End  floor  beams  are,  as  a 
rule,  used  in  the  case  of 
highway  bridges,  while  in 
the  case  of  railroad  bridges 
as  a  rule  they  are  omitted 
and  details  similar  to  those 
shown  in  Fig.  305  are  used, 
except  they  are  constructed 
to  fit  the  skew. 

The  outer  ends  of  the 
stringers  in  railroad  bridges 
should  be  square  with  the 
track,  as  indicated  in  Fig. 
394,  so  as  to  avoid  having 
ties  resting  upon  masonry  at 
one  end  and  upon  a  stringer 
at  the  other  end.  By  this 

Fig.  304  construction  we  also  obtain 

570 


SKEW  BRIDGES,  BRIDGES  ON  CURVES,  TRUSSES  AND  PORTALS      571 


equal  stringer  concentrations  on  the  first  intermediate  floor  beams,  as  is 
readily  seen. 

The  end  posts  in  each  end  panel  of  skew  truss  bridges  should  always 
be  parallel.  The  stresses  in  skew  bridges  are  determined  practically  in 
the  same  manner  as  previously  shown  for  square  spans;  the  only  differ- 
ence being  that  the  skew  must  be  taken  into  account,  which  is  done  by 
assuming  the  loads  to  be  applied  along  the  center  line  of  the  bridge.  Let 
the  diagram  at  (m)  in  Fig.  395  represent  the  plan  of  a  skew  truss  bridge. 
We  would  assume  the  live  load 
applied  along  the  center  line  gh. 
The  influence  line  for  the  reac- 
tion at  A  would  be  as  shown  at 
(o).  The  influence  lines  for  the 
stress  in  chord  cd  and  diagonal 
cD  would'  be  as  shown  at  (£) 
and  (v),  respectively.  To  sat- 
isfy the  criterion  for  maximum 
live-load  moment  about,  say, 
point  D,  the  distance  sg  would 
be  taken  as  A:  in  (5)  of  Art.  91 
and  as  gw  in  case  the  maximum 
moment  at  B  is  desired,  and  so 
on.  To  satisfy  the  criterion  (see 
Art.  90)  for  maximum  live-load 
shear  in,  say,  panel  EC,  the  load 
would  extend  from  h  to  and  into 
panel  BC  so  that  the  unit  load  in 
that  panel  (with  a  load  at  C) 
would  be  equal  (or  as  nearly 
equal  as  is  possible)  to  the  total 
load  on  the  span  divided  by  the 
distance  gh.  In  determining  the  Fig.  395 

dead-load  stress  the  panel  loads 

are  readily  computed  for  all  panel  points.  This  load  will  be  the  same  for 
all  panels  except  for  points  a,  /,  and  F,  in  which  case  a  slight  correction 
on  account  of  the  skew  must  be  made.  After  the  panel  loads  are  computed 
the  dead-load  stresses  are  readily  determined  either  analytically  or  graph- 
ically. However,  it  must  be  borne  in  mind  that  the  truss  is  unsym- 
metrical  with  reference  to  the  center  of  span. 

The  maximum  moment  and  shears  on  skew  through  plate  girder 
bridges  are  determined  in  exactly  the  same  manner  as  shown  above  for 
truss  spans.  Skew  deck  plate  girder  bridges  can  be  treated  the  same  as 
square  spans  without  any  appreciable  error. 

254.  Bridges  on  Curves. — Bridges  supporting  curved  tracks  are 
subjected  to  stresses  due  to  centrifugal  force  and  to  the  loads  being 
eccentrically  applied.  These  stresses  are  in  addition  to  dead,  live,  and 
impact  stresses. 

Stresses  Due  to  Centrifugal  Force  occur  in  the  laterals  system  at 
the  loaded  chord.  The  laterals  and  chords  constitute  the  system  which 
is  really  a  horizontal  truss.  For  centrifugal  force  in  general  we  have 


572 


STRUCTURAL  ENGINEERING 


F= 


Wv2 
gr 


(1) 


where  W  represents  the  weight  of  the  moving  body,  in  pounds,  v  the 
velocity  of  the  body  in  feet  per  second,  g  the  acceleration  due  to  gravity, 
and  r  the  radius  of  curvature  in  feet. 

Let  D  equal  degree  of  curvature.     Then  we  have 


50 


r  — 


and  hence  for  a  1°  curve  we  have 

50 


r  = 


-5,730ft. 


0.00873 
Let  V  equal  velocity  in  miles  per  hour.     Then  we  have 


= 


22 
3,600          15 


Now  substituting  these  values  in  (1)  we  have 
22"2 


F  = 


152x  32.2x5,730 


2  =  . 0000117  JFF2 (2) 


for  the  centrifugal  force  for  a  1°  curve  where  W  equals  weight  of 
the  moving  body  and  V  equals  the  velocity  in  miles  per  hour.  Therefore, 
for  a  4°  curve  and  50-mile  velocity,  we  have 


Then  if  W  be  the  uniform  live  load  per  ft.  of  span,  0.117  W  would 
be  the  centrifugal  force  per  ft.  of  span  which  would  be  used  to  deter- 
mine the  stress  in  the  chords  and  laterals.  As  an  illustration,  let  Fig. 


Fig.  396 


396  represent  the  plan  of  the  lateral  system  at  the  loaded  chord  of  a 
7-panel  bridge  supporting  a  track  on  a  4°  curve.  Let  W  represent  the 
live  load  per  ft.  of  span.  Then  for  a  velocity  of  50  miles  per  hour  we 
have  0.11 7W  for  the  centrifugal  force  per  ft.  of  span,  and  for  the  panel 
load  we  have 


Then  the  maximum  stress  in  the  laterals  and  chords  due  to  this  horizontal 


SKEW  BRIDGES,  BEIDGES  ON  CURVES,  TRUSSES  AND  PORTALS   573 


load  will  be  as  indicated  in  Fig.  396.  These  stresses  are  determined  in 
the  same  manner  as  previously  shown  for  wind  stresses.  For  example, 
to  obtain  the  maximum  stress  in  diagonal  eE  panel  points  B,  C,  and  D 
would  be  loaded  and  in  the  case  of  diagonal  fF  panel  points  E,  C,  D, 


Fig.  397 

and  E  would  be  loaded,  and  so  on.  The 
maximum  chord  stress  would  be  obtained  by 
loading  all  panel  points  from  B  to  G.  The 
compression  in  the  floor  beams  due  to  centrif- 
ugal force  is  not  considered,  as  the  laterals 
are  practically  always  connected  to  the  ten- 
sion flange  of  the  beams  and  consequently  the 
stress  due  to  centrifugal  force  only  tends  to  ° 
reverse  the  dead  and  live-load  tensile  stress 
in  those  flanges.  r 

The  stress  in  the  chords  due  to  centrif- 
ugal force  is  always  added  to  the  dead,  live, 
and  impact  stresses  in  summing  up  the  total 
maximum  stress. 

Stress  Due  to  Eccentricity  occurs  in  the  trusses,  stringers,  and  floor 
beams.  Let  the  diagram  shown  in  Fig.  397  represent  the  plan  of  a 
bridge  supporting  a  curved  track,  where  the  curve  acb  represents  the 
center  line  of  track  and  the  line  oo  represents  the  center  line  or  axis  of 
the  bridge  which  always  bisects  the  middle  ordinate  (d)  of  the  curve  as 
indicated.  It  is  readily  seen  that  loads  near  the  center  of  the  span  will 
load  the  outer  truss  and  stringer  more  than  the  inner  truss  and  stringer 
and  that  just  the  reverse  is  true  for  loads  hear  the  ends  of  the  span. 

The  diagram  in  Fig.  398  shows  the  conditions  at  or  near  the  center 
of  the  span  where  g  represents  the  center  of  gravity  of  the  load.  Taking 
moments  above  A  we  obtain 


Fig.  398 


r=W^      Fy 


(3) 


for  the  vertical  concentration  on  the  truss  at  B  where  W  represents  the 
weight  of  the  load  and  F  the  centrifugal  force.  If  the  super-elevation 
of  the  outer  rail  be  in  accord  with  the  speed  or  velocity  of  the  load  the 
resultant  of  W  and  F  will  pass  through  the  center  of  the  track,  in  which 
case,  taking  moments  about  A,  we  have 


for  concentration  on  the  outer  truss.  As  is  seen,  the  part  We/b  is  due 
to  the  eccentricity,  and  as  is  evident  this  is  greater  for  a  moving  load 
than  for  a  static  load,  for  in  the  latter  case  F  would  be  zero  and  e  would 
be  less,  if  not  in  reality  shifted  to  the  other  side  of  the  center  line  of  the 


574 


STRUCTURAL  ENGINEERING 


floor  beam.  If  the  velocity  of  the  load  be  greater  than  the  super-eleva- 
tion of  the  outer  rail  provides,  e  will  be  increased.  However,  such  a 
velocity  is  not  likely  to  occur  for  the  heavy  freight  trains  which  produce 
the  maximum  live-load  stress  in  the  trusses.  If  it  were  possible  to  locate 
accurately  the  center  of  gravity  (#)  of  the  load,  the  concentration  could 
be  determined  in  any  case  on  the  outer  truss  by  taking  moments  about 
A — which  would  be  simply  the  application  of  equation  (3) — and  like- 
wise the  concentration  on  the  inner  truss  could  be  determined  by  taking 
moments  about  B.  But  as  it  is  practically  impossible  to  locate  the  center 
of  gravity  of  the  load,  as  it  varies  so  widely  for  the  different  engines  and 
cars,  it  is  more  practical  to  consider  the  velocity  to  accord  with  the  super- 
elevation of  the  outer  rail  and  simply  consider  the  concentration  on  the 
outer  truss  to  be  as  expressed  by  equation  (4). 

As  is  seen  from  Fig.  397,  e  is  equal  to  one-half  of  the  middle  ordinate 
at  the  center  and  ends  of  the  span,  and  fair  results  will  be  obtained  by 
assuming  e  equal  to  one-half  of  the  middle  ordinate  throughout  the  entire 
length  of  the  span,  as  the  loads  near  the  center  of  span  contribute  the 
greater  part  of  the  chord  stresses  and  the  loads  near  the  end  contribute 
the  greater  part  of  the  web  stresses  and  hence  the  actual  discrepancy  is 
slight.  Then,  to  obtain  the  maximum  live-load  stresses  in  the  trusses, 
we  would  first  calculate  the  maximum  stress  in  each  member  the  same 
as  for  ordinary  bridges,  assuming  the  load  applied  along  the  center  line 
of  the  bridge,  and  then  increase  each  of  these  stresses  by  the  amount 
obtained  by  multiplying  the  stress  in  each  case  by  e/b  when  e  is  taken 
as  half  the  middle  ordinate  and  b  as  the  distance  between  trusses.  That 
is,  if  S  be  the  stress  in  a  member  due  the  load  considered  applied  along 
the  center  line  of  the  bridge,  the  stress  due  to  the  eccentricity  e  would  be 
S  x  e/b,  and  hence  the  total  stress  in  the  member  due  to  live  load  would 
beS  +  Se/b. 

In  case  more  accurate  results  be  desired  than  are  obtained  by  taking 
the  eccentricity  e  equal  to  one-half  the  middle  ordinate,  the  value  of  e 
at  each  floor  beam  can  be  computed  and  the  concentration  on  the  trusses 
at  each  of  those  points  determined  and  the  stresses  in  the  trusses  due  to 
same  computed.  However,  in  that  case  an  equivalent  uniform  live  load 
should  be  used,  as  the  work  would  be  quite  tedious  if  wheel  loads  be  used. 
When  the  stringers  are  symmetrically  placed  with  reference  to  the 
center  line  of  the  bridge,  the  maximum  moment  and  shear  on  the  outer 

stringers  are  obtained  by  first 

Truss  or  Mam  Girder-?  determining  the  moment  and 

shear  for  symmetrical  load 
and  then  increasing  each  of 
these  by  the  amount  obtained 
by  multiplying  each  by  e/b' 
where  b'  is  taken  as  the  dis- 
Fig-  309  tance  between  the  stringers. 

The  moment  and  shear  on  the 

inner  stringers  would  really  be  decreased  by  the  same  amount;  however, 
it  is  usual  practice  to  make  the  outer  and  inner  stringers  of  equal 
section. 

In  case  the  curve  is  quite  sharp  the  stringers  are  usually  off-set,  as 
shown  in  Fig.  399,  in  which  case  the  stringers  are  about  equally  loaded 


.----  — 

_  

51-ringerJ 

Truss  or  Main 


SKEW  BKIDGES,  BRIDGES  ON  CURVES,  TRUSSES  AND  PORTALS   575 

and  they  are  designed  as  in  ordinary  cases.  The  stringer  bracing  should 
be  designed  sufficiently  heavy  to  transmit  the  centrifugal  force.  The 
stress  in  this  bracing  is  determined  in  the  same  manner  as  shown  above 
for  the  lateral  system  of  the  structure. 

In  determining  the  maximum  moment  and  shear  on  the  floor  beams, 
the  eccentricity  of  the  loading  must  be  taken  into  account.  Otherwise 
the  work  is  the  same  as  previously  shown. 

In  the  case  of  double-track  bridges  the  determination  of  the  stresses 
due  to  centrifugal  force  is  the  same  as  for  single-track  structures,  except 
that  both  tracks  must  be  considered. 

According  to  the  A.R.E.  Specifications,  "structures  located  on  curves 
shall  be  designed  for  the  centrifugal  force  of  the  live  load  applied  at  the 
top  of  the  high  rail.  The  centrifugal  force  shall  be  considered  as  live 
load  and  derived  from  the  speed  in  miles  per  hour  given  by  the  expres- 
sion 60  -  2^Z),  where  D  equals  degree  of  curve." 

The  velocity  given  by  this  expression  (60  — 2JZ))  for  the  different 
degrees  of  curvature  'and  the  corresponding  centrifugal  force  are  as 
follows : 

D               V                  F  D                V                   F 

1° 57.5. 0.039JF  4° 50.0 0.117JF 

2° 55.0 0.071JF  5° 47.5 0.132JF 

3° 52,5 0.097JF  6°. 45.0 0.142ZF 

where  D  represents  the  degree  of  curvature,  V  the  velocity  (speed)  in 
miles  per  hour,  F  the  centrifugal  force  in  pounds,  and  W  the  weight  of 
the  load  in  pounds,  which  would  be  the  weight  of  load  per  ft.  of  span  if 
a  uniform  live  load  be  used. 

255.  Economic  Depth  of  Trusses. — As  is  evident,  a,  truss  is  of 
economic  depth  as  regards  weight  and  stiffeners  when  the  weight  of  the 
web  members  is  equal  to  the  weight  of  the  chord  members.  Of  course 
the  weight  of  the  floor  system,  which  varies  with  the  length  of  panels,  is 
part  of  the  total  weight  of  the  structure,  and  hence  the  panel  length  is 
involved.  There  are  so  many  variables  involved  that  the  theoretical 
determination  of  economic  heights  and  panel  lengths  is  practically  impos- 
sible. The  most  satisfactory  values  for  heights  and  panel  lengths  are 
obtained  by  actual  calculations.  The  following  values  are  recommended: 


Length  of  Span 
in  Feet 

-A1  or    tvaiiroau   jjriuges  — 

Length  of  Panels 
in  Feet 

Height  of  Truss 
in  Feet 

100 

25 

30 

125 

25 

30 

150 

25 

31 

175 

25 

33 

200 

25 

39 

225 

25 

45 

250 

25 

50 

300 

25 

55 

400   (14  panels) 
500   (16  Panels) 

28.58  (about) 
31.25 

62 

72 

Heights  for  intermediate  lengths  can  be  obtained  by  interpolating. 


576 


STEUCTUEAL  ENGINEEEING 


Length  of  Span 
in  Feet 
105 
135 
150 
162 
180 
200 


-For  Highway  Bridges- 
Length  of  Panel 
in  Feet 
15 
15 
16.6 
18 
20 
20 


Height  of  Truss 
in  Feet 
20 
22 
24 
28 
32 
36 


(a) 


^r3^,  /?* 

f^ 


In  order  to  obtain  the  required  overhead  clearance  and  a  satisfac- 
tory depth  of  portal  the  height  of  truss  is  limited  to  about  30  ft.  in  the 
case  of  railroad  bridges  and  to  about  20  ft.  in  the  case  of  highway 
bridges.  The  minimum  economic  panel  length  for  railroad  bridges  is 
about  25  ft.  and  the  maximum  about  33  ft.,  while  in  the  case  of  highway 
bridges  the  minimum  panel  length  is  about  15  ft.  and  the  maximum 
about  20  ft. 

256.  Economic  Span  Length. — The  shortest  bridge  in  the  case 
of  a  single  span,  as  a  rule,  is  the  cheapest,  as  the  cost  of  the  two  abut- 
ments in  most  cases  is  practically  independent  of  the  span  length.      How- 
ever, in  a  few  cases  the  profile  of  the  crossing  is  such 
that  the  cost  of  the  abutments  is  materially  affected 
by  their  location.     In  such  cases  care  must  be  taken 
in  locating  the   abutments   so  that  their   cost   is   a 
minimum,  the  usual  limit  being  when  the  cost  of  the 
span  equals  the  cost  of  the  two  abutments.    Often  the 
required  water  way  or  local  conditions  govern  the 
length  of  the  span,  in  which  cases  economy  of  span 
length  can  not  be  considered.     In  case  a  bridge  is 
composed  of  several  spans  the  spans  are  practically 
of  economic  length  when  the  cost  of  superstructure 
minus  the  floor  system  is  equal  to  the  cost  of  the  sub- 
structure minus  the  cost  of  the  end  abutments.     This 
is,  of  course,  assuming  that  the  total  length  of  the 
bridge  is  fixed  either  by  the  required  water  way  or 
local  conditions.    The  problem  can  be  solved  only  by 
trial.     First  a  profile  of  the  crossing  should  be  drawn 
carefully  to  scale  and  the  base  of  rail  or  roadway 
located  thereon,  and  next  the  cost  of  an  average  pier 
computed,  and  then  by  using  the  formulae  of  Art. 
124  (Art.  221  in  the  case  of  highway  bridges)  the 
weight. of  metal  in  the  spans  of  different  lengths  can 
be  obtained  (the  weight  of  the  floor  system  must  be 
subtracted  in  each  case)  and  by  careful  comparison 
of  the  different  bridges  possible  we  can  readily  as- 
certain the  economic  length  of  spans. 

257.  Stresses  in  Portals.— Portals  other  than  the  one  previously 
considered    (Art.    179)    are    sometimes    used.     The    plate    girder   portal 
shown  at  (a),  Fig.  400,  is  occasionally  used  in  case  of  bridges  having 
shallow  trusses. 


el  P+ 


(0 


Fig.   400 


SKEW  BRIDGES,  BEIDGES  ON  CURVES,  TRUSSES  AND  PORTALS      577 

Assuming  the  bottom  ends  of  the  end  posts  as  hinged,  the  wind 
pressure  tends  to  distort  the  end  posts  and  portal  as  indicated  at  (6). 

Considering  the  forces  to  the  right  of  section  SS  and  taking  moments 
about  d  [see  sketch  at  (c)  |  we  obtain 

F=2±M  ........  .........  ..(1) 

e  e 

for  the  stress  in  the  bottom  flange  of  the  portal. 

But  F=-R  +  P     and  H=R  +  P. 


Substituting  these  values  in  (1)  we  obtain 


When  x  =  0 


/v  c; 


F  =  0    .............................................  (3). 

When  x  —  w 

F  =  -^(fl  +  P)  ..........................  "  ..........  (4). 

Now  it  is  seen  from  the  above  equations  that  the  stress  in  the  bot- 
tom flange  of  the  portal  is  zero  at  the  center  of  the  portal  and  a  maximum 
at  each  end,  being  tension  at  one  end  and  compression  at  the  other. 

Taking  moments  about  b  we  obtain 


F  = 


for  the  stress  in  the  top  flange  of  the  portal.     Substituting 
for  V  and  ^(J2+P)  for  //  in  equation  (5)  we  obtain 


When«r  = 


When#  =  w/2 

F'-* 

~2 

When  x  —  w 


(R  +  P)+ (9). 


578 


STEUCTUBAL  ENGINEERING 


From  these  equations  it  is  seen  that  the  maximum  stress  in  the  top 
flange  of  the  portal  occurs  at  the  ends,  being  compression  at  one  end  and 
tension  at  the  other,,  and  that  the  minimum  stress  in  the  top  flange  occurs 
at  the  center  of  the  portal  where  it  is  equal  to  R/2. 

The  shear  on  the  portal  is  constant  throughout  its  length  and  is 
equal  to  V.  In  case  the  bottom  ends  of  the  end  posts  be  fixed  the  above 
will  apply  by  substituting  ^(h  —  e)  for  (h  —  e)  and  e  +  h/2  for  h  (see 
Art.  179). 

After  the  stresses  in  the  portal  are  determined  the  required  sections 
are  readily  computed.  Each  of  the  flanges  should  be  such  that  L/r  does 
not  exceed  120.  The  portal  shown  at  (a),  Fig.  401,  is  used  in  case  of 
limited  clearance. 

Assuming  the  bottom  ends  of  the  end  posts  to  be  hinged,  the  wind 
pressure  tends  to  distort  the  end  posts  and  the  portal  as  is  indicated 
at  (fc).  The  shear  between  g  and  /  is  constant  and  is  equal  to  V.  The 


Fig.    402 


moment  at  g  and  also  at  /  is  equal  to  Fx  b/2  which  is  obtained  by  taking 
moments  about  either  g  or  /.  The  moment  at  o  is  zero  and  at  any  point 
between  o  and  /  or  between  o  and  g  it  is  Vs.  Assuming  the  moment  to  be 
zero  at  a  and  k  the  strut  afgk  can  be  considered  as  a  continuous  beam. 

Then  taking  moments  at  either  /  or  g  we  have  (Vxb/2)  ~-d  for  the 
reaction  at  A:  or  a  which  is  the  shear  in  gk  or  af.  The  moment  at  any 
point  in  gk  or  af  is  equal  to  [(Fx  b/2)  -r  d~\x.  If  x-d  we  have  Vxb/2 
for  the  moment  at  g  or  /  which  is  the  same  as  found  above. 

The  horizontal  component  of  the  stress  in  knee  brace  gm  is  obtained 
by  considering  the  forces  to  the  right  of  section  2-2  and  taking  moments 
about  k.  Then  the  stress  in  gm  is  obtained  by  multiplying  this  com- 
ponent by  sec</>.  The  stress  in  knee  brace  cf  is  obtained  in  the  same  man- 
ner—considering the  forces  to  the  left  of  section  1-1. 

In  case  the  bottom  ends  of  the  end  posts  are  fixed  the  above  will 
apply  by  substituting  $(h-e)  for  (h-e)  and  e  +  \(h  —  e)  for  h. 

The  portal  shown  at  (a),  Fig.  402,  is  known  as  a  lattice  portal.  It 
is  distorted  by  the  wind  pressure  very  much  the  same  as  the  plate  girder 
portal. 

As  the  shear  is  constant  and  equal  to  V  throughout  the  length  of  the 
portal  the  diagonals  are  assumed  to  be  equally  stressed.  As  one  system 
is  in  compression  and  the  other  in  tension  the  moment  of  the  stresses  in 
the  diagonals  about  any  point  on  a  vertical  line  through  the  intersection  of 


SKEW  BEIDGES,  BBLDGES  ON  CUEVES,  TEUSSES  AND  POETALS   579 


the  diagonals  is  equal  to  zero  as  the  moments  balance.  Then  by  taking 
moments  about  c  and  considering  the  forces  to  the  left  of  any  section 
as  2-2,  shown  at  (6),  we  have 


T^  


Ilh-Vx 


for  the  stress  in  the  bottom  flange  at  b  and  taking  moments  about  b  we 
have 


H(h-e)-Vx      P 


for  the  stress  in  the  top  flange  at  c.     The  flange  stress  throughout  the 
portal  can  be  obtained  in  this  manner. 

The  stress  in  the  diagonals  is  obtained  by  dividing  the  shear  (F)  by 
the  number  of  diagonals  cut  by  a  vertical  plane,  and  multiplying  this  by 
the  secant  of  the  slope  of  the  diagonals. 

The  portal  shown  in  Fig.  403  is  used  to 
some  extent,  especially  for  long  span  bridges. 
Let  us  first  assume  that  the  diagonals  take  ten- 
sion only.  Then  diagonal  db  would  have  zero 
stress  when  the  wind  pressure  is  applied  as 
indicated  and  diagonal  ac  would  have  zero  stress 
if  the  wind  pressure  were  applied  from  the 
opposite  direction. 

Assuming  the  bottom  ends  of  the  end  posts 
hinged,  and  taking  moments  about  a  (ignoring 
diagonal  bd)  and  considering  the  forces  to  the 
left  of  section  2-2  we  obtain 


Fig.   403 


(i) 


for  stress  (compression)  in  be,  and  taking  moments  about  c  (ignoring 
diagonal  bd)  and  considering  the  forces  to  the  right  of  section  3-3  we 
obtain 

S>  =  *L,h-e)+(R+? 

6  \          2 


for  the  stress   (tension)   in  ad. 

The  stress  in  the  diagonal  ac  is  equal  to  Fsec<£.  If  the  wind 
pressure  were  applied  from  the  opposite  direction  to  that  indicated 
diagonal  bd  would  be  stressed,  and  the  stress  in  be  and  ad  would  be 
determined  by  taking  moments  about  d  and  b,  respectively.  The  stress 
in  diagonal  bd  would  be  Fsec</>  and  the  stress  in  diagonal  ac  would  be 


zero. 


In  case  the  diagonal  were  considered  to  take  compression  as  well  as 
tension  we  would  assume  the  two  equally  stressed  and  then  by  taking 
moments  about  o  and  m  the  stress  in  be  and  ad  could  be  determined,  as 
the  sum  of  the  moments  of  the  stresses  in  the  diagonals  about  these  points 
would  be  zero  as  they  would  balance. 


580 


STRUCTURAL  ENGINEERING 


If  the  lower  ends  of  the  end  posts  be  considered  fixed,  the  above 

would  apply  if  ^(h  —  e)  be  substituted  for  (h-e)  and  e  +  ^(h-e)   for  h. 

The  depth  of  the  portal,  as  a  rule,  is  so  shallow,  as  compared  with 

the  length  of  the  end  posts,  that  the  bending  of  the  posts  along  the  ends 

of  the  portal  can  be  ignored,  and  in  case  the  lower  ends  of  the  posts  are 

fixed  the  points  of  contra- 

ltt  /  flexure  in  the  posts  can  be 

N  ^*i  considered    as    being   mid- 

way between  the  portal 
and  the  lower  end  of  the 
posts,  as  has  been  done  in 
all  previous  examples.  But 
in  a  few  cases  the  required 
depth  of  the  portal  is  so 
great  that  the  bending  of 
the  end  posts  along  the 
ends  of  the  portal  must  be 
taken  into  account  in  de- 
termining the  points  of 
contra-flexure  in  the  posts. 
Let  the  diagram  shown, 

Fig-  404  (a),    Fig.    404,    represent 

such  a  case.  The  distor- 
tion of  the  portal  members  is  so  small  as  compared  with  the  flexure  of  the 
posts  that  the  points  C  and  B  can  be  assumed  to  remain  in  a  vertical  line 
and  likewise  the  points  D  and  E.  The  problem  is  to  determine  the  value 
of  z,  which  is  the  distance  from  the  lower  ends  of  the  end  posts  up  to  the 
points  of  contra-flexure. 

Let  R  represent  the  horizontal  component  of  the  forces  acting  upon 
the  post  at  C,  and  R'  the  horizontal  component  of  the  forces  at  B.  Then 
the  post  ABC  can  be  considered  as  a  beam  fixed  at  A  and  acted  upon  by 
the  forces  R  and  R '  as  indicated  at  (6). 

Then,  for  the  bending  moment  at  any  point  on  this  beam  between 
A  and  B  we  have 


(1). 


Integrating,  we  have 


dy 

But  -j*-  =  0  when  x  =  0,  therefore  C  =  0,  and  we  have 
ax 


(2), 


Integrating  again,  we  obtain 


SKEW  BRIDGES,  BRIDGES  ON  CURVES,  TRUSSES  AND  PORTALS      581 
But  y  —  0  when  #  =  0,  therefore  C'  —  0,  and  we  have 

(r2          r3\  /         r2         r3  \ 

"y-y)-K'(c|-T;  ................  «• 

For  the  bending  moment  at  any  point  between  C  and  B  we  have 

M'  =  EI^=R(h-x}  ...............................  (4). 

dx~ 

Integrating,  we  have 


As  is  obvious  (2)  and  (5)  are  equal  when  x-c  in  each.  So  by 
substituting  c  for  x  in  both  (2)  and  (5),  equating  and  reducing,  we 
obtain 


C"-     R'  °'' 
C   —R    - 

Substituting  this  value  of  C"  in  (5)  we  obtain 

2  \  2 

Of  O 

/C  /  76 

Integrating  again,  we  have 

^+C"' (6). 

As  is  obvious,  (3)  and  (6)  are  equal  when  x  =  c  in  each.     So  by  substi- 
tuting c  for  x  in  both   (3)   and   (G),  equating  and  reducing,  we  obtain 

C'"=+B'-j- 

Substituting  this  value  of  C' "  in  (6)  we  obtain 

fL.T4_£_7?'  fy\ 

X+        It     (( ). 


As  points  C  and  B  are  in  the  same  vertical  line  (3)  and  (7)  are 
equal  for  x  =  c  in  (3)  and  x  =  h  in  (7).  So  substituting  these  values, 
equating  and  reducing,  we  obtain 

R_  3c8  -  3c*fe       _  3c2 

^'  "  3/ic-2  -  c-3  -  2A3  ~  2hc  +  2h2  -  c2" 

When  x  —  z  we  have 


(9). 


From  this  equation  we  obtain 
R        (c-z) 


R'      (h-z) 


582  STRUCTURAL  ENGINEERING 

Now  equating  (8)  and  (9)  and  reducing  we  obtain 


By  applying  equation  (10)  the  points  of  contra-flexure  in  the  posts 
can  be  determined  and  then  the  stresses  in  the  portal  can  be  computed 
as  previously  explained. 


DRAWING  ROOM  EXERCISE  NO.  9 

(a)  Design  a  single-track  through  pin-connected  curve  chord  Pratt 
truss   railroad  bridge  and  make  a  complete  stress   sheet  of  same  upon 
tracing  cloth. 

Data: 

Length  of  span  =  9  panels  at  26'-0"  =  234'-0"  c.c.  end  pins. 

Height  of  truss  at  center  =  46'-0". 

Height  of  truss  at  hip  =  31'-0". 

Top  chord  joints  to  be  on  the  arc  of  a  parabola. 

Live  load,  Cooper's  E50. 

Dead  load,  to  be  assumed. 

Specifications,  A.  R.  E.  Ass'n. 

(b)  Make  a  general  detail  drawing  of  the  above  bridge. 

(c)  Design  a  single-track  through  Pettit  truss  railroad  bridge  and 
make  a  stress  sheet  for  same. 

Data: 

Length  of  span  =  14  panels  at  28'-0"  =  392'-0"  c.c.  end  pins. 
Live  load,  Cooper's  £50. 
Dead  load,  to  be  assumed. 
Specifications,  A.  R.  E.  Ass'n. 

(d)  Design  a  pony  truss  highway  bridge  and  make  a  general  draw- 
ing of  same  on  tracing  cloth. 

Data: 

Length  of  span  =  6  panels  at  8'-0"  =  48'-0". 

Height  of  truss  =  6'-6". 

Width  of  roadway  =  16'-0". 

Concrete  floor— 7"  slab  and  3"  gravel  fill. 

Specifications,  as  per  Art.  222. 

(e)  Design  a  through  pin-connected  curve  chord  highway  bridge. 

Data: 

Length  of  span  =  9  panels  at  18'-6"  =  166'-6"  c.c.  end  pins. 

Height  of  truss  at  center  =  28'-0". 

Height  of  truss  at  hip  =  20'-0". 

Roadway,  16'-0". 

Concrete  floor— 6"  slab  and  3"  gravel  fill. 

Specifications,  as  per  Art.  222. 


CHAPTER  XIV 

DESIGN  OF  BUILDINGS 
MILL  BUILDINGS 

258.  Preliminary. — The  simplest  type  of  mill  building  consists  of 
a  steel  roof  supported  upon  brick  or  concrete  walls  as  shown  at  (a), 
Fig.  405.  This  type  of  building  is  satisfactory  for  power  houses,  pumping 
stations,  and  small  manufacturing  plants  where  light  material  is  handled. 

The  type  shown  at  (b)  consists  of  a  steel  roof  supported  upon  steel 
columns.  The  side  and  end  walls  for  this  type  may  be  brick,  hollow 
tile,  concrete,  metal  lathing  plastered,  or  corrugated  iron.  In  case  the 
material  handled  is  heavy,  an  overhead  crane  would  be  installed.  This 
crane  would  run  lengthwise  of  the  building  and  be  supported  upon  girders 
riveted  to  the  columns. 

The  type  shown  at  (c)  is  practically  the  same  as  the  one  shown  at 
(6)  except  a  shed,  known  as  a  lean-to,  is  attached  to  each  side  and  a 
monitor  or  ventilator  is  built  onto  the  top  of  the  roof  for  the  purpose  of 
ventilating  and  lighting  the  building. 

The  type  shown  at  (d)  is  simply  a  double  building  of  the  same 
type  as  the  one  shown  at  (c)  except,  as  a  rule,  it  would  be  proportionately 
larger. 

The  type  shown  at  (e)  is  known  as  the  saw-tooth  construction.  This 
type  of  construction  is  used  in  the  case  of  a  very  wide  building  in  order 
to  obtain  light  in  the  interior  of  the  building. 

All  of  the  types  shown  above  are  for  steep  roof  construction  where 
the  roof  covering  is  corrugated  iron,  slate,  concrete,  or  tile.  In  case  of 
flat  roof  construction,  where  the  roof  covering  is  felt  covered  with  tar 
and  gravel,  the  types  shown  in  Fig.  406  are  used. 

The  type  shown  at  (/)  consists  of  a  steel  roof  supported  upon  brick 
or  concrete  walls. 

The  type  shown  at  (g)  consists  of  a  steel  roof  supported  upon  steel 
columns.  The  end  and  side  walls  may  be  corrugated  iron,  brick,  con- 
crete, hollow  tile,  or  metal  lathing  plastered. 

The  type  shown  at  (h)  is  the  same  as  the  one  shown  at  (g)  except 
a  lean-to  is  added  to  each  side. 

The  type  shown  at  (Ar)  is  practically  the  same  as  the  one  shown  at 
(h)  except  each  lean-to  is  wider  and  the  building  is  proportionately 
larger  in  every  way. 

The  types  of  buildings  shown  in  Figs.  405  and  406  are  general 
types.  These  are  often  modified  to  suit  certain  special  requirements; 
in  fact  it  is  impossible  to  indicate  types  suitable  for  all  cases. 

As  a  rule  the  nature  of  the  material  to  be  handled  and  the  required 
location  of  the  machinery  govern  the  type  of  building  used. 

583 


Purlin 


Monitor  or  l/enti lator^^. 


Windows  or  Louvresy/r^. 


Mf?C/OW5-~ 

x-^v 

|Z«7/7-/O 

fe»r-f 

g  ^x 

/      «,      \ 

vv^ 

I 
"^•Column 

1 

I 
"^-Column 

584 


(9) 


Clear 


Lean-fo 


Glass 


(jlass 

^AlA 

7_ 

\ 

7^^^ 

^  G/ass 

-ffi3=F*q 

7 

\ 

cr^r^rrr-rr, 

^j/\y\ 

/\ 

i  

^i't* 

C-  { 

7\P\l7^^ 

dCS> 

cf&> 

'     I 

I 

\    ' 

Fig.  406 


585 


586 


DESIGN  OF  BUILDINGS 


587 


The  maximum  length  of  span  for  the  trusses  as  a  rule  is  limited  by 
the  length  of  the  cranes  which  have  a  maximum  length  of  about  80  ft. — 
more  often  40  or  50  ft.  is  used. 

The  names  of  the  different  parts  of  an  ordinary  steel  mill  building 
are  given  in  Fig.  407.  These  same  names,  however,  apply  in  general 
to  steel  mill  buildings. 

The  most  common  forms  of  roof  trusses  are  shown  in  Fig.  408.  The 
steep  pitch  roofs  are  used  when  the  roof  covering  is  corrugated  iron, 
slate,  or  tile,  and  the  flat  roofs  are  used  when  the  roof  covering  is  felt 


(d)    Fink  Truss 


(b)  Fan  Truss. 


(  CJ  Warren  Truss . 


(d)  Pratt  Truss. 


(e)  Flat-  Warren  Truss. 


(f)  Flat  Praff  Truss. 


Ho  we  Truss . 


(h)  Flat  HoweTruss. 


Fig.   408 


covered  with  tar  and  gravel.  Concrete  roof  covering  may  be  used  in 
either  case. 

The  details  shown  in  Fig.  409  are  those  of  a  truss  supported  upon 
columns.  The  purlins  are  channels  which  are  supported  on  the  roof 
truss  at  panel  points.  The  roof  covering  is  corrugated  iron  connected  to 
the  purlins  by  steel  bands. 

The  details  shown  in  Fig.  410  are  those  of  a  roof  truss  supported 
upon  walls.  The  roof  covering  is  slate  laid  on  2"  sheathing.  The 
purlins  are  channels  which  have  6"  x  3"  nailing  strips  bolted  to  them. 
In  this  case  the  purlins  are  not  at  panel  points  and  the  top  chord  must 
be  designed  heavy  enough  to  resist,  in  addition  to  the  direct  stress,  the 
cross  bending  caused  by  the  purlin  loads.  Wooden  purlins  as  shown 
at  (a)  are  sometimes  used,  which  is  very  good  construction. 

The  details  shown  in  Fig.  411  are  those  of  a  roof  without  purlins. 
The  sheathing  consists  of  2"x4"  timbers  laid  on  edge,  acting  as  beams 
extending  from  truss  to  truss.  The  bay  length  in  the  case  of  such  con- 
struction should  be  limited  to  about  14  or  15  ft.  In  case  the  roof  load 


588 


STRUCTURAL  ENGINEERING 


on  the  trusses  is  quite  heavy  and  not  supported  at  panel  points,  the  top 
chords  of  the  trusses  are  constructed  of  two  angles  and  a  plate  as  shown 
in  Fig.  411.  The  plate  is  principally  for  transmitting  shear. 


Fig.  409 

The  details  shown  in  Fig.  412  are  those  of  an  ordinary  monitor 
where  louvre  bars  are  used.  The  details  shown  in  Fig.  413  are  those  of 
an  ordinary  monitor  where  pivoted  windows  are  used. 

Pitch  of  a  roof  truss  is  the  ratio  of  its  height  to  its  length.  Thus: 
A  roof  truss  60  ft.  long  and  20  ft.  high  would  have  J  pitch,  and  if  its 
height  were  30  ft.  the  pitch  would  be  J.  The  terms  "pitch"  and  "slope" 


Fig.  410 


590 


{STRUCTURAL  ENGINEERING 


Fig.    412 


are  somewhat  confusing.  The  pitch  of  an  ordinary  truss  is  obtained  by 
dividing  its  rise  or  height  by  its  total  length,  while  the  slope  is  obtained 
by  dividing  its  height  by  one-half  of  its  length. 


Fig.   413 


A  lean-to  truss,  as  shown  at  (d),  Fig.  405,  can  be  considered  as  a 
half  truss. 

The  pitch  of  a  roof  depends  upon  the  roof  covering.  The  following 
is  recommended: 


DESIGN  OF  BUILDINGS  591 

Roof  Covering  Pitch 

Corrugated    Iron not  less  than  J. 

Slate not  less  than  J,  preferably  J  or  J. 

Tile not  less  than  £. 

Felt,  tar,  and  gravel.  .  .not  more  than  ^5,  preferably^ 

259.  Dead  Load. — Except  for  the  roof  trusses,  the  dead  load  for 
mill  buildings  is  readily  estimated.      In  making  these  estimates  the  fol- 
lowing units  of  weights  may  be  used : 

Timber    4.0  Ibs.  per  sq.  ft.  B.  M. 

Concrete     150.0  Ibs.  per  cu.  ft. 

Corrugated  iron.,  #16 3. 5  Ibs.  per  sq.  ft. 

Corrugated  iron,,  #18 2.7  Ibs.  per  sq.  ft. 

Corrugated  iron,  #20 1.9  Ibs.  per  sq.  ft. 

Corrugated  iron,  #22 1.5  Ibs.  per  sq.  ft. 

Slate,  J"  thick,  3"  double  lap...  4.5  Ibs.  per  sq.  ft. 

Slate,  fV  thick,  3"  double  lap.  .  .  6.5  Ibs.  per  sq.  ft. 

Tile    (plain) 18.0  Ibs.  per  sq.  ft. 

Tile   (Spanish) 8.5  Ibs.  per  sq.  ft. 

Glass,  y  thick 2.0  Ibs.  per  sq.  ft. 

Gravel,  felt,  and  tar  roof  covering.  5.0  Ibs.  per  sq.  ft. 

Hollow  tile   (gross) 40.0  Ibs.  per  cu.  ft. 

The  approximate  weight  of  roof  trusses  in  pounds  per  sq.  ft.  of  roof 
surface  is  as  follows: 

—Pitch- 
Flat  i  i  J  i 

20-ft.  span 2.0  1.9  1.8  1.7  1.5 

30-ft.   span 2.6  2.5  2.3  2.1  1.8 

40-ft.   span 3.3  3.2  2.9  2.8  2.4 

50-ft.  span 3.8  3.6  3.4  3.2  2.7 

60-ft.  span 4.6  4.4  4.1  3.8  3.2 

70-ft.   span 5.2  4.9  4.6  4.3  3.7 

80-ft.   span 5.7  5.4  5.1  4.8  4.1 

90-ft.  span 6.4  6.1  5.7  5.3  4.6 

100- ft.   span 7.0  6.7  6.2  5.8  5.0 

120-ft.   span 8.0  7.6  7.1  6.7  5.7 

260.  Snow  Load. — The  snow  load  on  roofs  varies  with  the  locality, 
slope  of  roof,  and  kind  of  roof  covering.     According  to  Trautwine,  fresh 
fallen  snow  weighs  from  5  to  12  Ibs.  per  cu.  ft.,  while  moistened  snow 
compacted  by  rain  will  weigh  from  15  to  50  Ibs.  per  cu.  ft. 

The  following  loads  in  pounds  per  sq.  ft.  of  roof  surface  are  con- 
sidered to  be  sufficient  allowance  for  snow: 

Pitch 

Locality  Flat       J         J         i         *         £ 

Central   States 30       30       25       20       15       10 

Northwestern  States 40       40       38        30       20        12 

New  England  States 36       36       32       27       18       12 

Rocky    Mountain    States 35       35       30       26       15       10 


592  STRUCTURAL  ENGINEERING 

These  loads  should  be  reduced  20  per  cent  in  case  the  roof  covering 
be  slate  or  corrugated  iron. 

No  snow  load  need  be  considered  south  of  latitude  35°.  However, 
in  some  localities  south  of  this  latitude  a  load  of  10  Ibs.  per  sq.  ft.  of 
roof  surface  should  be  included  in  roof  loads  to  allow  for  sleet.  In  fact, 
load  due  to  sleet,  in  all  cases,  should  be  taken  as  10  Ibs.  per  sq.  .ft.  of  roof 
surface. 

261.  Wind  Load. — The  wind  pressure  against  vertical  surfaces 
varies  with  the  velocity  of  the  wind  and  to  some  extent  with  the  area  of 
the  exposed  surface.  The  wind  load  (pressure)  on  vertical  surfaces  can 
be  safely  considered  as  30  Ibs.  per  sq.  ft.  except  for  steel  towers,  where 
the  exposed  surfaces  are  small,  in  which  case  the  load  should  be  con- 
sidered as  40  Ibs.  per  sq.  ft.  This  30  Ibs.  pressure,  in  the  case  of  mill 
buildings,  is  considered  to  be  applied  to  the  vertical  sides  and  ends  of 
the  buildings. 

In  the  case  of  wind  load  on  sloping  roofs  the  pressure  is  assumed  to 
act  perpendicularly  or  normally  to  the  roof  surface.  This  pressure  can  be 
determined  only  experimentally.  Hutton  derived,  from  his  experiments 
in  1788,  the  following  formula: 

r>       T>  s  -     \ 1-842  cosct  ~  * 
Pn  =  P  (sma) 

for  the  normal  pressure  due  to  wind  upon  inclined  surfaces,  where  Pn 
represents  the  normal  pressure,  P  the  corresponding  horizontal  pressure, 
and  a  the  slope  of  the  surface. 

Duchemin,  in  1829,  derived  in  the  same  manner  the  following 
formula : 


=  p/    2  sina    \ 
\l  +  sin2a/ 


for  the  normal  pressure  on  inclined  surfaces,  where  the  letters  signify  the 
same  as  in  Hutton's  formula. 

These  two  formulas  are  in  general  use  and  the  values  for  the  normal 
pressures  when  P  equals  30  Ibs.  are  as  follows: 

Normal  Pressure 
Slope  of  Roof  Hutton         Duchemin 

5°    3.9                    5.1 

10°    7.2  10.2 

15°    10.5  14.2 

18°-26'   (i    pitch) 13.0  17.4 

21°-48'    ( J  pitch) 15.0  19.8 

26°-34'   (i   pitch) 18.0  22.4 

30°   19.9  24.0 

33°-41'  (£    pitch) 22.0  25.5 

40°    25.1  26.7 

45°    (i   pitch) 27.1  28.2 

60°   30.0  30.0 

262.     Cranes. —  In  designing  a  mill  building,  it  is  absolutely  neces- 
sary that  the  designer  have  complete  data  as  regards  the  type,  weight, 


DESIGN  OF  BUILDINGS 


593 


and  capacity  of  the  cranes  to  be  used,  and  also  the  clearance  required 
for  their  operation. 

The  data  given  in  Fig.  414  will  suffice  in  most  cases  for  overhead 
cranes.      However,   the    general    requirements    in    each    case   should    be 


Span 


Q 


Truck 


Capacity 

Span 

ft 

<q' 

B 

c 

EdchlVhee 

Total  Net  W 
of  Crane 

5  Tons 

4-0  Ft 

A-'-ll' 

5-  7" 

0'-9" 

e'-e" 

IZ800 

22400 

5     •• 

(SO  .- 

5-3" 

5'   ll" 

o^7^ 

6'-  S" 

15500 

31300 

1  O  • 

AO  - 

5-10 

6-  6 

o-  /o' 

9-  2" 

19800 

264QO 

1  O    • 

60  - 

e'-o' 

6'-S" 

o'-/o" 

9-  2" 

22700 

37600 

1  5  - 

-4O   •• 

6-0" 

6-9' 

o-  10' 

9  -  2" 

26500 

33900 

1  5  •• 

6O  - 

6-2" 

6-n" 

0-  10" 

9-2" 

29600 

44000 

20  •• 

^4O    • 

6-2' 

7-  3" 

o-n" 

9-8" 

32700 

37600 

2O  •• 

60   •• 

6'-  5" 

7-6" 

O-  ll" 

9-8" 

37000 

50700 

25   •• 

^.o  • 

6-5" 

71  7" 

o-n" 

9-6" 

39300 

4340O 

25 

6O  - 

6-9" 

a'-o" 

O'-M" 

10-10" 

44500 

58700 

25  - 

SO   •• 

7-2" 

S-  5" 

O'-M" 

13-4." 

50800 

7990O 

3O  .. 

-40  •• 

e'-a" 

S'-  0" 

-0" 

\O'-A" 

46200 

495OO 

30  - 

6O    • 

v-o" 

SL  6' 

'-o" 

II-  2' 

51700 

66600 

3O    • 

ao  - 

7L5" 

8-  /J" 

'-O" 

13-4' 

56800 

9O7OO 

4-O  •• 

-40    - 

r1  o" 

6L9" 

-o" 

ll'-4" 

601  00 

64200 

4-O  •• 

60    - 

-?-*• 

9  -  l" 

'-o" 

n'-io" 

67000 

64300 

4-O  - 

SO    •• 

7-9" 

9-6' 

-o" 

13-4" 

7560O 

II  29OO 

50  •• 

^.0   •• 

7-7" 

9'-  5" 

-o" 

II  '-4" 

74000 

76100 

5O  •• 

<ZG  •• 

7-9" 

9-7" 

-o" 

IIL0"  J:/0' 

^IZOO 

9850O 

50  •• 

&0   - 

a-^" 

.0'-3" 

'-o" 

O1?'  5-/' 

>*590O 

131  200 

Fig.  414 

carefully  studied;  and  often  it  is  advisable  to  consult  the  company  that 
manufactures  the  crane  to  be  used. 

Complete  Design  of  a  Roof  Supported  upon  Brick  Walls 

263.    Data.— 

Size  of  building  =  80  ft.  long  by  40  ft.  wide. 
Length  of  roof  trusses  =  40'-0"  c.c.  end  bearings. 
Height  of  roof  trusses  =  lO'-O"   (J  pitch). 
Length  of  bay  =  20'-0". 


594 


STEUOTUEAL  ENGINEEEING 


Roof  covering,  slate  laid  on  2"  sheathing. 
Purlins,  steel  channels. 

264.  Design  of  Purlins. — As  2"  sheathing  is  specified,  the  first 
thing  to  do  is  to  determine  the  distance  that  the  purlins  can  be  apart,  so 
that  the  sheathing  will  not  be  overstressed  or  the  deflection  be  too  great. 

The  load  on  the  sheathing  includes  snow,  wind,  and  the  weight  of 
the  slate  and  sheathing. 

The  maximum  snow  and  wind  load  is  not  likely  to  occur  at  the  same 
time  as  the  snow  would  be  blown  off.  Let  us  assume  that  the  wind  pres- 
sure is  18  Ibs.  (Hutton — see  Art.  261)  and  the  snow  10  Ibs.  per  sq.  ft. 
of  roof  surface.  The  10  Ibs.  snow  load  provides  for  sleet,  ice,  or  frozen 
snow.  Using  the  weights  given  in  Art.  259,  we  have  the  following  dead 
load  per  sq.  ft.  of  roof: 

iV'   slate        =    6.5  Ibs.  per  sq.  ft.  of  roof  surface 

2"  sheathing  =    8.0  Ibs.  per  sq.  ft.  of  roof  surface 

14.5  Ibs.  per  sq.  ft.  of  roof  surface 

Adding  this  to  the  snow  load  we  obtain  24.5  Ibs.  This  load  acts 
vertically  while  the  wind  load  acts  normally  to  the  roof. 

Let  <f>  be  the  slope  of  the  roof.  Then  for  the  normal  pressure  due 
to  dead  and  snow  load  we  have 

24.5  x  cos0  =  24.5  x  0.895  =  21.9,  say  22  Ibs. 
Now  adding  this  to  the  wind  load  we  have 

18  +  22  =  40  Ibs. 

for  the  total  maximum  normal 
pressure  on  the  roof  per  sq.  ft. 
of  roof  surface.  Let  x  be  the 
distance  between  purlins,  as  in- 
dicated in  Fig.  415,  and  consider- 
ing the  sheathing  to  be  a  simple 
beam  1  ft.  wide,  we  have 


\ 


M  =  i f  x  40  x  x2  x  12  =  60#2  inch  Ibs. 

for  the  maximum  bending  mo- 
ment. Taking  1,000  Ibs.  as  the 
allowable  unit  stress  on  the 
sheathing  and  applying  Formula 
(C)  of  Art.  53,  we  have 


Fig.    418 

from  which  we  obtain 


=  1,000  x  8, 


*=11.5  ft. 


for  the  distance  that  the  purlins  could  be  apart  and  the  sheathing  would 
not  be  overstressed  from  the  40  Ibs.  load.  The  sheathing  should  be 
capable  of  supporting  a  200-lb.  man  in  addition  to  the  12.5  Ibs.  dead 
load.  Considering  the  man  to  be  midway  between  purlins  (center  of 


DESIGN  OF  BUILDINGS  595 

span)  and  resolving  the  loads  normally  to  the  roof,  we  have 

M'  =  J  (14.5  x  0.895  x  x2. x  12)  +  —  x  0.895  x  -  x  12  =  19.4o?2  +  537,r 

for  the  bending  moment.     Then  we.  have 

19.4*2  +  537^  =  1,000x8, 
from  which  we  obtain 

*  =  10.8  ft. 

for  the  distance  between  the  purlins  which  is  only  0.7  ft.  less  than  found 
above  for  the  40  Ibs.  load.  But  we  know  from  experience  that  such 
spacing  would  not.  be  practical  as  the  deflection  of  the  sheathing  would 
be  so  great  that  the  slate  would  be  broken.  From  this  it  is  seen  that 
the  spacing  of  the  purlins  depends  entirely  upon  the  maximum  allowable 
deflection  of  the  sheathing.  This  deflection  should  not  exceed  ^J-^-  of 
the  distance  between  th'e  purlins.  For  the  deflection  due  to  the  200-lb. 
man  at  mid  span  (see  Art.  65)  we  have  the  following  expression: 


Placing  this  equal  to  ,r/800  and  taking  E  as  1,000,000  and  JP  as  200,  we 
have 

200  x  0.895  x  ,r3  _    x 

48  x  l,000,00(Xx  8  ~  800' 
from  which  we  obtain 

*  =  51.8"  or  about  4'-4". 

If  the  deflection  due  to  the  dead  load  of  14.5  Ibs.  per  sq.  ft.  of  roof 
surface  be  taken  into  account  the  value  of  *  will  be  a  little  less  than  4'-4", 
so  the  correct  distance  will 
be  about  4'-0".  The  actual 
spacing  of  the  purlins  de- 
pends to  some  extent  upon  ^*^*\/  '  ^ 
the  design  of  the  roof  truss.  ^sj\  si^  /  i 
In  this  case  a  Fink  truss  <^/\^  / 
will  be  used  and  the  most 
suitable  spacing  obtainable 
of  the  purlins  is  shown  in 
Fig.  416.  This  gives  a  dis- 
tance of  about  3 '-8"  be- 
tween purlins.  Fig.  410 

The    dead     and     snow 

load  as  given  above  is  24.5  Ibs.  per  sq.  ft.  of  roof  surface.  Adding  to 
this  the  wind  load  of  18  Ibs.,  which  we  will  consider  to  act  the  same 
as  the  dead  load  on  the  purlins,  we  have 

24.5  +  18  =  42.5,  say  43  Ibs. 

for  the  total  load  per  sq.  ft.  of  roof  surface. 

Now,  as  the  purlins  are  3'-8"  apart,  we  have 

43x3.66  =  157.38.  say  158  Ibs, 


596 


STRUCTURAL  ENGINEERING 


for  the  roof  load  per  ft.  of  purlin.     Assuming  the  purlin  to  weigh  12  Ibs. 
per  ft.  we  have 

158 +  12  =  170  Ibs. 


for  the  total  load  per  ft.  of  purlin, 
moment  on  the  purlin  we  have 


Then  for  the  maximum  bending 


M'  =  Jx  170x20   x  12  =  102,000  inch  Ibs. 

Dividing  this  by  16,000  we  obtain  a  section  modulus  of  6.3  which 
calls  for  a  1"  x  12J#  [  or  an  8"  x  11J*  [.  The  8"  x  11  J#  [  will  be  used 
for  purlins  throughout. 

Theoretically  the  slope  of  a  purlin  should  be  taken  into  account 
in  determining  the  maximum  stress  on  it,  but  as  the  above  result  would 
not  be  changed  by  so  doing  (especially  if  the  stiffening  effect  of  the 
nailing  strips  and  sheathing  be  taken  into  account)  we  are  really  justified 
in  ignoring  the  slope,  as  was  done  in  the  above  calculations. 

265.  Determination  of  Stresses  in  the  Trusses  Due  to  Dead 
and  Snow  Load. — There  are  really  two  loads  to  consider:  The  dead 
and  snow  loads  that  can  be  combined  with  the  maximum  wind  load,  and 
the  dead  and  maximum  snow  load  that  could  be  combined  with  one-third 
of  the  maximum  wind  load.  The  first  is  given  above  as  24.5  Ibs.  per  sq. 
ft.  of  roof  surface  where  the  snow  (sleet)  is  taken  as  10  Ibs.  Combining 
this  with  the  maximum  wind  load  (Hutton)  we  obtained  the,  43  Ibs.  given 
above.  For  the  other  load,  taking  the  maximum  snow  for  the  Central 
States,  we  have 


Snow 
20 


Dead  Load 
14.5        + 


Wind 
6     =40.5  Ibs. 


which  is  less  than  the  combination  first  considered.  So  we  will  use  the 
24.5  Ibs.  dead  and  snow  load.  The  end  reaction  on  each  purlin,  except 
the  eave  and  ridge  purlins,  including  the  weight  of  truss  and  purlins,  is 
[(24.5  +  2.9)  3.66+11.25]  10  =  i;il6  Ibs.  about,  and  hence  the  concentra- 
tion on  the  truss  at  each  of  these  purlins  is  1,116x2  =  2,232  Ibs.  Then 
the  load  on  the  truss  will  be  as  shown  in  Fig.  417. 


4-0-0' 


Fig.   417 


The  concentration  on  the  panel  points,  considering  the  top  chord  in 
each  panel  as  a  simple  beam,  is  shown  at  (a),  Fig.  418.  The  diagram  of 
the  stresses  is  shown  at  (b).  This  diagram  is  constructed  by  beginning 
at  joint  A  and  passing  around  each. joint  clock-wise.  Thus,  laying  off  1-2 


DESIGN  OF  BUILDINGS 


597 


equal  to  the  reaction  and  2-3  equal  to  the  1,480  Ibs.  load  and  closing  on 
1  and  3  we  obtain  1-2-3-4-1  for  the  stress  diagram  for  joint  A.  Starting 
at  4  and  passing  around  joint  a  clock- wise,  we  obtain  4-3-5-6-4  for  the 


-/8500  * V*  -16500  *  \yf  -ft 000  * 


(b) 


Fig.   418 

stress  diagram  for  that  joint.  Starting  at  1  and  passing  around  joint  B 
clock- wise  we  obtain  1-4-6-7-1  for  the  stress  diagram  for  that  joint. 
The  stress  diagrams  for  joints  b  and  C  can  not  be  drawn  as  the  structure 
stands  as  there  are  three  unknown  forces  at  each  joint.  We  get  around 
this  difficulty  by  inserting  the  temporary  member  Cc  and  omitting  mem- 
bers be  and  ce.  Having  made  this  modification  we  obtain  the  stress 
diagram  7-6-5-9-8-7  for  joint  b.  Beginning  at  1  and  passing  around 
joint  C  clock- wise  we  obtain  1-7-8-12-*-!  for  the  stress  diagram  for  that 
joint.  The  only  correct  stress  obtained  by  drawing  the  last  two  diagrams 
is  the  stress  in  CD  which  is  represented  by  the  line  *-l.  This  stress, 
as  is  readily  seen,  is  not  affected  by  the  changing  of  the  web  members. 
Having  determined  the  stress  in  CD  we  can  consider  members  be  and  ce 
in  place  and  Cc  omitted,  and  proceed  with  the  analysis.  Considering 
joint  C  and  passing  around  clock- wise  we  obtain  s-l-7-13-s  for  the  stress 
diagram  for  that  joint.  Having  the  stress  in  bC  determined,  which  is 
represented  by  the  line  7-13,  we  can  draw  the  stress  diagram  for  joint  b 
which  is  13-7-6-5-9-10-13. 

The  remainder  of  the  diagram  at  (b)  is  readily  constructed.  This 
diagram  represents  the  stresses  in  the  left  half  of  the  truss.  The  stresses 
in  the  right  half  are  the  same,  for  corresponding  members,  and  hence 
the  diagram  at  (6)  is  sufficient  for  determining  the  stress  in  all  the  truss 
members.  By  scaling  the  diagram  at  (6)  the  desired  stresses  in  the 
truss  members  are  obtained.  These  are  shown  on  the  members  at  (a). 


598 


STRUCTURAL  ENGINEERING 


266.  Determination  of  the  Stresses  in  the  Trusses  Due  to  Wind 
Load.— The  wind  load  (see  table  of  Art.  2G1)  is  18  Ibs.  (Hutton)  per 
sq.  ft.  of  roof  surface.  The  combined  dead  and  snow  load  considered 
in  the  last  article  is  30.5  Ibs.  per  sq.  ft.  of  roof  surface.  So  by  multiply- 
ing the  loads  shown  in  Fig.  418  by  ^085  we  obtain  the  wind  loads 
shown  at  (a),  Fig.  419,  which,  of  course,  act  normally  to  the  roof  and 
upon  one  side  only. 

The  two  ends  of  the  trusses  are  considered  to  be  equally  fixed.  Then 
the  reactions  will  be  parallel  to  the  loads.  Laying  off  the  load  line  MN, 


(b) 


Fig.  419 

shown  at  (b),  and  drawing  the  ray  diagram  MON,  and  constructing  the 
corresponding  equilibrium  polygon  ab'  .  .  .0,  and  drawing  the  line  OE 
parallel  to  the  closing  line  ag,  we  obtain  ME,' which  represents  the  reaction 
R  at  A,  and  EN,  which  represents  the  reaction  R\  at  F. 

Then  beginning  at  joint  A  and  passing  around  each  joint  clock-wise, 
as  explained  in  the  last  article,  the  stress* diagram  shown  at  (6)  is  readily 
drawn.  By  scaling  this  diagram  the  desired  stresses,  which  are  given  at 
(a),  are  obtained. 

267.  Designing  of  the  Truss  Members.— Combining  the  stresses 
shown  in  Figs.  418  and  419  we  obtain  the  stresses  shown  in  Fig.  420, 
which  are  maximum  stresses,  and  which  the  truss  members  must  be 
designed  to  transmit. 

The  members  will  be  designed  so  that  }"  rivets  can  be  used  through- 
out, which  requires  that  the  flanges  of  the  angles  through  which  rivets 
pass  be  not  less  than  2£". 


DESIGN  OF  BUILDINGS 


599 


Bottom  Chord  AB.  This  is  a  tension  member  and  the  required  net 
area  of  cross-section  is  equal  to  28,340  -f-  16,000  =  1.77n".  Two  angles 
will  make  the  most  satisfactory  section  for  this  member.  The  least  size 
permissible  is  2|"  x  2|",  as  }"  rivets  are  to  be  used,  and  besides  they 


Fig.  420 

should  be  at  least  this  size  to  secure  the  necessary  rigidity.  So  we  will 
use  2— Ls  2J"  x  2£"  xj"  =  2.38-  0.42  =  1.96n",  which  has  0.19n"  more 
section  than  required,  but  the  thickness  is  a  minimum  and  hence  this 
section  is  as  near  the  required  as  is  possible  to  obtain.  This  same  section 
will  be  used  for  the  bottom  chord  throughout,  as  it  is  the  minimum. 

Members  cJc  and  cB.  These  are  tension  members  and  the  required 
section  is  only  about  0.25n",  so  we  will  use  1— L  2J"  x  2"  xj"  =  1.06- 
0.21  =  0.85°",  which  is  the  smallest  angle  permissible  if  f"  rivets  are 
used. 

Members  ek  and  kC.  These  are  tension  members  and  the  required 
section  for  ek  is  0.9n",  and  for  JcC  it  is  0.61n".  We  will  use  2— Ls  2£"  x 
2"  x^r/  =  1.7n//  net,  which  are  the  smallest  angles  permissible. 

Member  cC.  This  is  a  compression  member  which  has  a  length  of 
67".  The  value  of  L/r  should  not  exceed  120.  Then  for  the  allow- 
able r  we  have  67/120  =  0.55.  Let  us  try  2— Ls  2£"x  2"xi"  =  2.12n". 
If  the  2"  flanges  are  turned  out  the  least  radius  of  gyration  will  be  0.79. 
Then  we  have 

16,000  -  70  x  -^-  =  10,060  Ibs. 
\j  •  i  y 

for  the  allowable  unit  stress.     Dividing  this  into  the  stress  we  obtain 
8,750  -r  10,060  =  0.87  sq.  ins. 

for  the  required  area  of  cross-section,  which  is  about  one-third  of  the 
area  contained  in  the  above  angles,  but  these  angles  will  be  used  as  they 
are  as  small  as  we  can  use.  The  L/r  =  67/0.79  =  0.85,  which  is  compara- 
tively low. 


600  STRUCTURAL  ENGINEERING 

Members  dk  and  bB.  We  will  make  each  of  these  members  of  2  —  Ls 
2J"  x  2"  x  J",  which  are  minimum  size  angles,  the  same  as  used  for  cC. 
The  actual  requirement,  however,  can  be  ascertained  in  the  same  manner 
as  shown  above  for  cC  but,  as  is  obvious,  this  work  is  not  necessary. 

Top  Chord.  This  is  a  compression  member.  Considered  in  the 
vertical  direction  the  length  would  be  a  panel  length,  and  considered  in 
the  transverse  direction  the  length  should  be  taken  as  two  panel  lengths, 
owing  to  there  being  some  question  as  to  the  actual  support  given  to  the 
chord  by  the  purlins. 

The  length,  considering  the  transverse  direction,  is  from  A  to  c, 
which  is  about  ll'-2"  or  134".  Owing  to  the  purlin  concentration  being 
applied  between  panel  points  there  will  be  cross  bending  on  the  chord 
which  it  must  resist  in  addition  to  the  direct  stress.  Let  us  assume 
2—  Ls  5"x3J"x-&"  =  7.06n".  The  5"  legs  will  be  placed  vertically  to 
resist  cross  bending.  Assuming  that  the  angles  are  f"  apart,  (&-&),  the 
least  radius  of  gyration,  is  1.50.  Then  we  have 


16,000  -  70  x    ~r  =  9,750  Ibs. 
1.50 

for  the  allowable  unit  stress  in  compression.     The  concentrations  on  the 
top  chord  due  to  dead  and  snow  load  are  given  in  Fig.  417.     The  chord 

in  each  panel  can  be  considered  as  a  fixed 
beam.  The  maximum  moment  will  occur 
at  supports  b  and  d.  The  loads  in  panel 
Ab  (the  same  as  in  the  other  panels)  are 
shown  in  Fig.  421.  This  load  (at  g)  is 
obtained  by  resolving  the  1,800  Ibs.  purlin 
concentration  due  to  dead  and  snow  load 
(see  Fig.  417)  normally  to  the  roof  and 
adding  the  concentration  due  to  the  18  Ibs. 
wind  load,  which  is  equal  to  18x3.66x 
Fig.  421  20  =  1,317  Ibs.  Hence,  for  the  concentra- 

tion at  g  we  have 

1,800x0.895  +  18x3.66x20  =  2,928,  say  2,900  Ibs. 
Applying  (7)   of  Art.  69,  we  have 

M'  =  2,900x67  (2P-P-fc)  .............................  (1) 

for  the  maximum  moment  which  occurs  at  6. 


Substituting  this  value  of  k  in  (1),  we  obtain 
M'  =  2,900  x  67  (2  x  0.102  -  0.033  -  0.32)  =  28,215,  say  28,000  inch  Ibs. 

for  the  maximum  bending  moment  on  the  top  chord. 

The  moment  of  inertia  of  the  2  —  Ls  5"  x  3£"  x  Ty  about  the  gravity 
axis  perpendicular  to  the  long  legs  is  8.9  x  2  =  17.8,  and  the  distance  from 
the  top  of  the  chord  down  to  this  same  axis  is  1.63.  Then  applying 


601 


602 


STRUCTURAL  ENGINEERING 


C  of  Art  53,  we  have 


M'  =  31,000  = 


/x20 


from  which  we  obtain 

/  =  5,mibs. 

for  the  maximum  compressive  unit  stress  due  to  cross  bending      For  the 
maximum  unit  stress  due  to  direct  compression,  we  have 

35,000  -r8n"  =  4,375  Ibs. 

Now  adding  this  to  the  stress  due  to  cross  bending,  found  above,  we  have 

1,375  =  Ibs.   9,550   (about). 


for  the  actual  maximum  unit  stress  on  the  top  chord.  The  allowable, 
as  found  above,  is  9,750  Ibs.  So  the  2—Ls  5"  x  3J"  x  J"  =  8n",  assumed 
above,  are  about  the  correct  size,  and  hence  will  be  used. 

This  completes  the  necessary  calculations  and  next  the  details  can 
be  drawn.  Complete  shop  drawings  for  the  building  are  shown  in  Fig. 
422.  The  details  shown  here  are  considered  to  be  self-explanatory. 
However,  the  student  should  verify  the  riveting. 


Fig.  423 


268.    Determination  of  Stresses  in  Trusses  Supported  upon 
Columns. — Stresses  due  to  dead  and  snow  load  are  determined  just  the 


DESIGN  OF  BUILDINGS 


603 


same  as  if  the  trusses  were  supported  upon  walls,  and  are  fully  treated 
in  Art.  265.      The  knee  braces  are  ignored  in  the  analysis. 

Stresses  Due  to  Wind  Load.     The  columns  are  fixed  to  the  masonry 
by  anchor  bolts,  which  should  extend  well  into  the  masonry,  and  by  the 


Fig.   424 


addition  of  knee  braces  the  columns  are  quite  firmly  connected  to  the 
trusses,  so  that  the  columns  can  really  be  considered  as  fixed,  in  which 
case  the  wind  pressure  will  cause  them  to  bend  as  indicated  in  Fig.  423, 
where  0  and  0'  are  the  points  of  contra-flexure. 

In  determining  the  stresses  in  the  truss  due  to  wind  pressure,  we 


604  STRUCTURAL  ENGINEERING 

first  locate  the  points  of  contra-flexure  in  the  columns  by  applying  (10) 
of  Art.  257.  After  the  points  of  contra-flexure  are  located,  we  next 
consider  the  ground  moved  up  to  that  level  as  indicated  at  (#)  in  Fig.  424. 

We  next  compute  the  value  of  the  wind  loads,  PI,  P2,  etc.  In  com- 
puting the  horizontal  loads  (PI  and  P2)  we  take  the  wind  pressure  at 
30  Ibs.  per  sq.  ft.  of  vertical  surface,  and  in  computing  the  value  of  the 
loads  (P3  .  .  .  P7)  normal  to  the  roof  we  use  the  pressure  per  sq.  ft. 
of  roof  surface  given  in  Art.  261,  the  actual  intensity  of  the  pressure 
used,  of  course,  being  in  accord  with  the  slope  of  the  roof. 

After  the  intensities  of  the  loads  (PI  .  .  .  P7)  are  computed,  we 
next  determine  the  horizontal  and  vertical  components  of  the  reactions 
on  the  columns  at  O  and  0'.  If  the  columns  are  of  equal  moment  of 
inertia  we  assume  that  the  horizontal  components  are  equal  and  that  each 
is  equal  to  one-half  of  the  horizontal  component  of  all  of  the  loads. 
These  horizontal  components  can  be  graphically  determined  by  laying 
off  the  loads  as  shown  at  (fo),  drawing  the  vertical  AP,  and  the  horizontal 
PB.  Then  the  horizontal  component  in  each  case  is  represented  by  one- 
half  of  the  line  PB.  Let  H  represent  this  component.  To  determine 
the  vertical  components  of  the  reactions  at  O  and  0'  we  complete  the  ray 
diagram  at  (fr),  then  construct  the  corresponding  equilibrium  polygon 
z-n-o  .  .  .  y  as  shown  at  (a).  Then  by  prolonging  the  segments  zn  and 
uy  to  intersect  at  I  and  from  I  drawing  a  line  parallel  to  the  line  AB, 
which  represents  the  resultant  F  of  all  of  the  loads,  we  obtain  the  line 
of  action  of  the  resultant,  and  prolonging  this  resultant  to  intersection  at 
N  with  line  00',  and  then  laying  off  O'M  equal  to  the  vertical  component 
(=AP)  of  all  of  the  loads,  and  drawing  the  line  MO  and  the  ordinates 
NT  and  TU,  we  have  the  vertical  component  F2  of  the  reaction  at  O' 
represented  by  the  line  NT  and  the  vertical  component  V\  of  the  reaction 
at  0  by  the  line  TU. 

The  line  MO  is  really  an  influence  line  for  the  vertical  component 
of  reaction  at  O' '.  To  show  that  this  is  true,  let  us  assume  that  the  re- 
sultant F  passes  through  point  0' .  Then,  as  is  readily  seen,  the  total 
vertical  component  of  all  of  the  loads  would  be  taken  by  column  kO' ',  in 
which  case  the  vertical  component  on  column  Oc  would  be  zero,  and  if 
the  resultant  F  intersected  the  line  OO'  midway  between  the  two  columns 
the  vertical  components  of  the  reactions  at  0  and  Of  would  be  equal  and 
each  equal  to  one-half  of  the  ordinate  O'M.  So,  evidently,  the  line  MO 
is  the  influence  line  for  the  vertical  component  of  the  reaction  at  0' '. 
That  is,  if  the  resultant  F  intersects  the  line  00'  at  any  point  C  between 
0  and  O'  the  vertical  component  of  the  reaction  at  O'  and  0  are  repre- 
sented by  the  ordinate  CD  and  DG,  respectively. 

If  the  resultant  F  intersects  the  line  00'  at  any  point  between  O 
and  0'  the  vertical  components  of  the  reactions  will  be  positive;  but  if  it 
intersects  at  any  point  E  to  the  right  of  0'  the  vertical  component  at  0 
would  be  negative,  and  the  vertical  component  at  0'  positive.  In  that 
case  the  ordinate  EK  (MO  prolonged  to  K)  would  represent  the  positive 
vertical  component  of  the  reaction  at  0',  while  MS  would  represent  the 
negative  vertical  component  of  the  reaction  at  0.  This  last  construction 
is  readily  proven:  Taking  moments  about  0  we  have 


DESIGN  OF  BUILDINGS  605 

from  which  we  obtain 


But  from  similar  triangles  we  have 

OE__EK_     KK 

00'  ~  O'M  ~"    V  ' 

from  which  we  obtain 

EK=(FxOE)  +  00', 
and  therefore 

EK=F2. 

After  the  horizontal  and  vertical  components  of  the  reactions  at  0 
and  Of  are  thus  determined,  the  actual  resultant  reaction  at  each  of  the 
points  is  readily  determined,  but  it  is  really  just  as  convenient  to  use  the 
components  in  determining  the  stresses  in  the  truss. 

Having  determined  the  horizontal  and  vertical  components  of  the 
reactions  at  0  and  0'  in  the  above  manner,  we  can  readily  determine 
graphically  the  stresses  in  the  truss  due  to  the  wind  loads  by  first  drawing 
the  auxiliary  frames  Oac  and  O'wk  and  beginning  at  either  0  or  0'  and 
passing  around  each  joint  in  the  same  direction.  The  stress  diagram  at 
(c)  is  obtained  by  starting  at  0  and  passing  around  each  joint  clock- 
wise. The  stresses  obtained  in  the  auxiliary  trusses,  including  the 
stresses  in  the  columns,  are  to  be  ignored,  but  the  stresses  in  the  knee 
braces  and  truss  proper  are  not  affected  by  adding  the  auxiliary  trusses 
and  hence  they  are  correct  stresses. 

Buildings  with  Lean-tos.  In  case  the  lean-tos  are  of  simple  con- 
struction, as  indicated  at  (/),  Fig.  425,  each  lean-to  column  should  be 
considered  as  acting  only  as  a  simple  beam  in  resisting  the  wind  pressure, 
in  which  case  the  wind  pressure  on  the  side  of  a  lean-to  would  be  trans- 
mitted equally  to  the  top  and  bottom  of  the  columns.  Then,  considering 
the  case  shown  in  Fig.  425,  the  pressure  on  the  side  AB  would  be  trans- 
mitted equally  to  points  A  and  B,  producing  the  loads  PI  and  P2  (each  = 
3,000  Ibs.)-  The  horizontal  pressure  of  the  loads  from  B  to  C  would  be 
transmitted  directly  to  the  column  at  C.  The  horizontal  pressure  repre- 
sented as  F  (=8,600  Ibs.)  is  obtained  by  constructing  the  force  diagram 
shown  at  (6).  Part  of  the  force  F  is  transmitted  to  point  D  and  part 
to  E.  The  results  obtained  will  be  accurate  enough  if  we  consider  the 
part  DE  of  the  main  column  as  a  simple  beam.  Then  taking  moments 
about  E  we  have 


for  the  part  of  F  transmitted  to  point  D.  Then  adding  this  force  /  to 
the  3,200  Ibs.  at  D,  we  obtain  the  9,650  Ibs.  force  shown  at  (d).  The 
part  of  F  transmitted  to  D  can  also  be  determined  by  constructing  the 
influence  line  shown  at  (c). 

After  the  force  or  load  at  D  (all  other  loads  are  assumed  as  known) 
is  determined,  the  ray  diagram  shown  at  (e)  can  be  drawn  and  the  cor- 
responding equilibrium  polygon  a-b-c-d-e-f-g-h  ...  w  at  (d)  constructed. 
Then  prolonging  Im  and  ab  to  intersection  at  I  and  drawing  from  I  a 


Fig.  425 


606 


DESIGN  OF  BUILDINGS 


607 


line  parallel  to  the  resultant  GK  we  obtain  10 "  for  the  line  of  action  of 
the  resultant  of  all  the  wind  loads  affecting  the  roof. 

By  laying  off  O'P  (=F  =  8,400)  equal  to  the  vertical  component  of 
all  of  these  loads,  and  drawing  ON  and  NO",  we  have  the  vertical  com- 
ponent F'2  of  the  reaction  on  the  column  at  0'  represented  by  the  dis- 
tance NO" ,  and  the  vertical  component  V\  of  the  reaction  on  the  column 
at  0  represented  by  the  distance  MP.  F2  is  positive  and  V\  negative. 
Each  of  the  horizontal  components  H  on  each  column  is  equal  to  one-half 
of  the  distance  SK. 

Now,  having  determined  all  of  the  loads  and  the  horizontal  and 
vertical  components  of  the  reactions  at  O  and  0'  we  can  determine  the 
stresses  in  the  roof  truss  in  exactly  the  same  manner  as  shown  above  for 
the  building  without  lean-tos. 

In  case  the  lean-tos  are  of  the  construction  shown  in  Fig.  426,  the 
wind  pressure  will  cause  the  columns  to  bend  as  indicated.  If  the  shear, 
indicated  as  HI,  H2,  etc.,  and  the  direct  stress  on  the  columns  at  the 


Fig.   426 

points  of  contra-flexure — that  is,  the  horizontal  and  vertical  components 
of  the  reactions  at  these  points — were  determined,  there  would  be  no 
difficulty  in  determining  the  stresses  in  the  main  and  lean-to  trusses. 
The  actual  values  of  the  horizontal  and  vertical  components  of  the  reac- 
tions at  the  points  of  contra-flexure  depend  not  only  upon  the  relative 
stiffness  of  the  columns  but  upon  the  stiffness  of  the  trusses  as  well.  But 
as  this  relative  stiffness  cannot  be  determined  at  all  until  the  structure 
is  fully  designed,  and  then  only  to  a  limited  extent,  it  is  evident  that  the 
method  used  for  determining  the  wind  stresses  must  necessarily  be  based 
to  some  extent  upon  practical  assumptions. 

By  considering  the  main  truss  and  the  parts  of  the  columns  above 
the  points  of  contra-flexure,  0  and  O',  as  an  independent  structure,  shown 
at  (a),  Fig.  427,  we  can  determine  the  horizontal  and  vertical  components 
of  the  reactions  at  0  and  O'  by  drawing  the  ray  diagram  at  (b)  and  the 
corresponding  equilibrium  polygon  at  (a)  and  the  influence  line  OJ,  all 
of  which  has  been  previously  explained. 

Considering  the  lean-to  and  the  part  of  the  main  column  below  the 


608 


STRUCTURAL  ENGINEERING 


point  of  contra-flexure  O  as  an  independent  structure,,  shown  at  (c),  we 
know  H'  and  V  which  are  applied  at  O,  as  they  are  equal  and  opposite, 
respectively,  to  the  //'  and  V  found  at  (a),  and  we  also  know  the  wind 
loads,  but  the  components  of  the  reactions  at  the  points  of  contra-flexure 


Fig.   427 


S  and  £'  and  likewise  the  force  H"  are  unknown.  The  force  H"  is  the 
reaction  from  the  part  of  the  structure  above  0  due  to  the  wind  loads  on 
the  lean-to,  and  hence  is  equal  to  the  part  of  these  loads  that  is  trans- 
mitted through  the  main  truss  to  the  other  side  of  the  building.  The 
actual  value  of  H"  depends  upon  the  rigidity  of  the  columns  and  trusses. 
This  rigidity  can  be  ascertained  only  by  tedious  analysis  of  the  structure 
after  it  is  designed.  However,  a  sufficiently  accurate  value  of  H"  can 
be  obtained  from  the  following  empirical  formula: 


(i) 


where  77  represents  the  horizontal  component  of  the  wind  loads  on  the 
lean-to  and  L  the  length  of  the  main  roof  truss  and  e  the  distance  from 


DESIGN  OF  BUILDINGS 


609 


the  ground  to  the  top  of  the  lean-to  and  b  the  distance  from  the  top  of 
the  lean-to  to  the  knee  brace  (see  Fig.  426). 

The  values  of  e,  b,  and  L  are  known  (always)  and  H  can  be  deter- 
mined by  constructing  the  force  polygon  CDE  shown  at  (e).  By  sub- 
stituting in  (1)  H"  can  be  determined  and  then  constructing  the  ray 
diagram  at  (e),  and  drawing  the  corresponding  equilibrium  polygon  at 
(c)  and  the  influence  line  ST,  we  obtain  the  vertical  components  V\  and 
V%  of  the  reactions  at  S  and  S',  and  then  the  only  unknown  forces  re- 
maining are  the  horizontal  components  of  these  reactions  which  are  rep- 
resented at  H3  and  H4.  These  forces  (H3  and  H4)  would  be  consid- 
ered to  be  equal  if  the  two  columns  were  of  equal  moment  of  inertia,  but 
as  they  never  are  we  must  assume  relative  values.  We  will  assume  the 
lean-to  column  to  be  one-third  as  rigid  as  the  main  column.  Then  as 
,  we  have 


where  H0  represents  the  total  horizontal  component  of  all  the  forces  on 
the  structure  shown  at  (c),  including  those  at  0. 

Considering  the  lean-to  and  part  of  the  main  column  shown  at  (d) 
as  an  independent  structure,  the  only  applied  forces  are  H"  ,  V"  ,  and  H' 
at  0'  '.  These  forces  having  been  previously  determined,  we  can  deter- 
mine the  vertical  component  of  the  reactions  at  the  points  of  contra- 


V4 


Fig.  428 


flexure,  S"  and  S"  ',  by  constructing  the  force  polygon  at  (/)  and  draw- 
ing from  0'  the  line  O'Y  parallel  to  the  resultant  LM  and  then  the 
influence  line  S"Z. 

Assuming  the  rigidity  of  the  lean-to  column  to  be  one-third  of  that 
of  the  main  column,  we  have 

H6  =  -Hf0andH5  =  ^Hf0. 

4  4 

We  now  have  all  of  the  components  of  the  reactions  at  the  points 


610  STRUCTURAL  ENGINEERING 

of  contra-flexure  determined,  and  we  can  now  proceed  with  the  determi- 
nation of  the  stresses  in  the  structure. 

By  considering  the  part  of  the  structure  above  the  points  of  contra- 
flexure  0  and  0'  and  adding  the  auxiliary  trusses  to  the  sides  as  indicated 
at  (tz),  Fig.  428,  the  stresses  in  knee  braces  and  main  truss  can  be  graph- 
ically determined  very  readily  by  beginning  at  either  0  or  Of  and  passing 
in  the  same  direction  around  each  joint. 

By  considering  the  part  of  the  structure  at  (c),  Fig.  428,  as  an 
independent  structure  and  adding  the  auxiliary  frames,  shown  dotted, 
the  stresses  in  the  knee  brace  and  lean-to  truss  can  be  graphically  deter- 
mined very  readily  by  beginning  at  S  and  passing  in  the  same  direction 
around  each  joint.  Likewise  the  stresses  in  the  knee  brace  and  the 
lean-to  truss  shown  at  (d),  Fig.  428,  can  be  determined. 

269.  Determination  of  the  Stresses  in  the  Columns. — The  col- 
umns are  subjected  to  direct  stress  and  cross  bending  similar  to  the  case 
of  the  end  posts  in  bridges.      (See  Art.  180.) 

After  the  horizontal  components  of  the  reactions  at  the  points  of 
contra-flexure  are  determined,  the  bending  moment  at  any  point  in  a 
column  is  readily  computed.  For  example,  the  bending  moment  at  any 
point  in  column  DE  (Fig.  425)  is  equal  to  Hx,  being  a  maximum  at 
D  and  E. 

Complete  Design  of  a  Mill  Building 

270.  Data.— 

Nature  of  building,  machine  shop. 

Length  of  building=ll  bays  @  20'-0"  =  220'-0". 

Width  of  main  shop  =  60'-0"  c.c.  of  main  columns. 

Width  of  lean-tos  =  24'-0"   (lean-to  on  each  side  of  building). 

Roof  covering,  No.  20  corrugated  iron. 

Allowable  intensities  on  metal — 

Tension 16,000  Ibs.  per  sq.  in. 

Compression.  .16,000-70   L/r  where  L  is  the  length   of 
member  in  ins.  and  r  the  least  radius  of  gyration  of 
the  cross-section  in  ins. 
Crane 20-ton  overhead  in  main  shop. 

271.  Preliminary. — In  starting  the  designing  the  very  first  thing 
to  do  is  to  draw  a  sketch  of  the  cross-section  of  the  building  as  shown  in 
Fig.  429,  showing  the  general  requirements  as  to  type  and  pitch  of  roof, 
spacing  of  purlins,  height  of  lean-to,  clearance  for  crane,  height  of  clear 
story,  etc.     The  roof  covering  is  to  be  of  corrugated  iron.     The  pitch  of 
the  main  roof  will  be  ^  and  the  lean-tos  about  £  (see  Art.  258).     The 
maximum  spacing  of  the  purlins  is  governed  by  the  kind  of  roof  covering. 
In  this  case  No.  20  corrugated  iron  will  be  used  and,  as  stated  in  the 
Carnegie  handbook,  the  maximum  span  must  not  be  more  than  6  ft.,  and 
less  is  preferable.      In  this  case  the  purlins  will  be  spaced  about  4'-6" 
centers.    The  distance  from  the  knee  brace  to  the  crane  girder,  as  specified 
in  Fig.  414,  is  6'-2". 

272.  Designing  of  the  Purlins.— The  weight  of  the   corrugated 
iron,  as  per  Art.  259,  is  1.9  Ibs.  per  sq.  ft.  of  roof  surface.     The  weight 


DESIGN  OF  BUILDINGS 


611 


of  the  purlin  will  be  assumed  to  be  12  Ibs.  per  ft.  of  length.  The  snow 
load  (frozen  snow  and  ice)  will  be  taken  as  10  Ibs.  per  sq.  ft.  of  roof 
surface  and  the  wind  load  as  22  Ibs.  per  sq.  ft.  of  roof  surface  as  per 


Fig.   429 

Art.  261  for  roof  having  ^  pitch.  Then  for  the  total  load  on  each  inter- 
mediate purlin  per  ft.  of  length  we  have  the  following: 

Dead  load  =  1.9x4.5  +  12=  20.5\r~~ 
Snow  load  =  10  x4.5  =  45.0  J  °D'D 
Wind  load  =  22  x4.5  =  99.0 

164.5,  say  165  Ibs. 

Then  for  the  maximum  moment  on  the  purlin  we  have 

M  =  J  x  165  x  202  x  12  =  99,000  inch   Ibs. 

Dividing  this  moment  by  16,000  we  obtain  99,000  +  16,000  =  6.2  for  the 
section  modulus,  which  calls  for  a  7"xl2.25#  channel  or  an  8"xll.25# 
channel.  Use  1 — [  8"  x  11.25*  for  each  purlin. 

273.  Designing  of  the  Lean-to  Rafters. — The  wind  load  acts 
perpendicularly  to  the  rafter  while  the  dead  and  snow  load  acts  vertically. 
So,  for  the  concentration  on  the  rafter  at  each  intermediate  purlin,  using 
the  same  unit  loads  as  used  in  the  last  article,  except  the  wind  load  which 
is  taken  as  18  Ibs.  (the  lean-to  roof  has  i  pitch),  we  have: 

Dead  load  =  20.5  x  20  x  0.908=  372  Ibs. 
Snow  load  =  45.0x20x0.908=  817  Ibs. 
Wind  load  =  (18  x  4.5)  x  20  =1,620  Ibs. 

2,809  Ibs.,  say  2,800  Ibs. 

Then  applying  this  concentration  at  each  purlin  and  considering 
the  rafter  as  a  simple  beam  we  have  the  case  fully  represented  in  Fig. 
430.  The  maximum  moment  on  the  rafter  due  to  these  loads,  as  is 
obvious,  will  occur  at  C,  the  center  of  span.  So  taking  moments  about  C 
we  have 


612 


STRUCTURAL  ENGINEERING 


M  =  [7,000  x  13.2  -  (2,800  x  8.8)  -  2,800  x  4.4]  12  =  665,280  inch  Ibs. 

for  the  maximum  moment  on  the  rafter  due  to  concentrated  loads.     As- 
suming the  rafter  to  weigh  42  Ibs.  per  ft.  of  length  we  have 


M'  =  Jx42x26.4   x  12  =  43,900  inch  Ibs. 


i 
o 

| 

fc 

* 

c    ^J       §3| 

<3.d 

44' 

i 

?' 

13.2' 

264' 

Adding 


Fig.   430 

for   maximum   moment   on   the   rafter   due   to   its   own   weight, 
together  the  above  moments  we  have 

665,280  +  43,900-709,180  inch  Ibs. 

for  the  total  maximum  moment  on  the  lean-to  rafter.  Dividing  this  by 
16,000  we  obtain  44.3  for  the  section  modulus  which  calls  for  a  12"x40# 
or  15"x42#  I.  We  will  use  the  15"x42#  I  for  each  lean-to  rafter. 

274.    Determination  of  Stresses  in  Trusses  Due  to  Snow  and 
Dead  Load. — The  snow  and  dead  load  per  ft.  of  purlin  as  given  in  Art. 

272  is  65.5  Ibs.  Then  for  the 
concentration  on  the  truss  at 
each  intermediate  purlin  we 
have  65.5x20  =  1,310  Ibs. 
Then,  as  two  of  these  con- 
centrations are  transmitted 
to  each  panel  point  of  the  top 
chord  of  the  truss  (see  Fig. 
429),  we  obtain  1,310x2  = 
2,620  Ibs.  for  the  part  of  the 
panel  load  due  to  snow,  roof 
covering,  and  purlins.  The 
weight  of  the  roof  truss  as 
per  Art.  259  is  3.8  Ibs.  per 
sq.  ft.  of  roof  surface.  Then, 
for  the  part  of  the  panel  load 
due  to  the  weight  of  the  roof 
truss  we  have  3.8x9x20  = 
684  Ibs.  Now,  adding  this 
to  the  part  due  to  the  roof 
load  we  have  2,620  +  684  = 
3,304,  say  3,300  Ibs.  (1,800 
Ibs.  of  this  is  due  to  snow) 
for  the  panel  load  due  to 
snow  and  dead  load,  and  hence  the  snow  and  dead  load  on  the  truss  will 
be  as  shown  at  (a),  Fig.  431.  By  beginning  at  A  and  constructing  the 


Fig.   431 


DESIGN  OF  BUILDINGS 


613 


stress  diagram  shown  at  (6)  the  stresses  shown  on  the  truss  are  obtained. 
Multiplying  each  of  these  stresses  by  1,500/3,300  the  stresses  marked  D 
which  are  due  to  dead  load  alone  are  obtained. 

275.  Determination  of  Stresses  in  Trusses  Due  to  Wind  Load. 
—The  wind  load  per  sq.  ft.  of  roof  surface  is  22  Ibs.  (Hutton),  as  per 
Art.  261.  Then  for  the  panel  load  on  the  roof  we  have  22  x  9  x  20  =  3,960 
Ibs.  and  hence  the  wind  loads  on  the  truss  will  be  as  shown  in  Fig.  432. 


Fig.   432 

The  horizontal  load  at  a  is  equal  to  30  x  5  x  20  =  3,000  Ibs.  The  load  at  b 
will  be  the  same  as  at  a  plus  the  horizontal  component  of  the  wind  load 
from  the  lean-to.  The  horizontal  load  at  the  top  and  bottom  of  the 
lean-to  column  is  equal  to  30x7x20  =  4,200  Ibs.,  as  shown  at  (6).  The 
pitch  of  the  lean-to  roof  is  about  J  (i|),  so  the  wind  pressure  per 
sq.  ft.  of  roof  is  18  Ibs.,  as  per  Art.  261.  Then  the  concentration  on  the 
rafter  at  each  intermediate  purlin  is  equal  to  18x4.5x20  =  1,620  Ibs. 
and  hence  the  wind  loads  on  the  lean-to  roof  will  be  as  shown  at  (6). 


614  STRUCTURAL  ENGINEERING 

Then  by  constructing  the  force  polygon  at   (c)   we  obtain 

F  =  8,250  Ibs. 

for  the  horizontal  component  of  the  forces  on  the  lean-to.  Then  adding 
this  to  the  30x5x20  =  3,000  Ibs.  from  the  clear  story  we  obtain  3,000  + 
8,250  =  11,250  Ibs.  for  the  total  wind  load  at  b.  Now  having  all  the 
wind  loads  determined  we  can  proceed  with  the  determination  of  the 
stresses  in  the  knee  braces  and  truss  due  to  same. 

The  points  of  contra-flexure  in  each  column  will  be  considered  as 
being  midway  between  the  bottom  of  the  column  and  knee  brace.  By 
constructing  the  ray  diagram  shown  at  (<^),  and  the  corresponding  equi- 
librium polygon  c-d-e  .  .  .  n-m  at  (a),  and  prolonging  the  segments  cd 
and  nm  to  intersection  at  /,  and  drawing  from  I  a  line  parallel  to  the 
resultant  AB,  we  obtain  the  line  of  action  of  the  resultant  of  all  of  the 
forces,  which  intersects  the  horizontal  line  00'  at  N. 

Then  laying  off  0' S  equal  to  the  vertical  component  of  all  of  the 
forces  and  drawing  the  influence  line  OS  and  dropping  a  vertical  from 
N  we  have  the  vertical  component  of  the  reaction  at  0'  given  by  the 
ordinate  NT  and  the  vertical  component  of  the  reaction  at  0  by  the  ordi- 
nate  TU.  The  horizontal  components  J/l  and  H2  of  the  reactions  at  0' 
and  0  are  equal  and  each  is  equal  to  one-half  of  the  horizontal  component 
of  all  the  forces,  which  is  represented  at  (e?)  by  the  line  CB. 

Now  having  the  components  of  the  reactions  at  the  points  of  contra- 
flexure,  O  and  0',  determined,  the  stresses  in  the  knee  braces  and  truss 
due  to  wind  loads,  which  are  given  at  (a),  are  determined  by  adding  the 
auxiliary  trusses,  shown  dotted,  and  constructing  the  stress  diagram 
shown  at  (e).  The  diagram  shown  at  (e)  is  obtained  by  beginning  at  O 
and  passing  around  each  joint  clock-wise.  The  diagram  at  (e)  should 
be  verified  by  the  student. 

276.  Designing  of  the  Truss  Members. — The  maximum  stresses 
in  each  truss  member  (as  given  in  Figs.  431  and  432),  and  also  the  ap- 
proximate lengths  of  the  members,  are  shown  on  the  one  diagram  in  Fig. 
433.  Having  this  data  we  can  proceed  with  the  designing  of  the  members. 

Member  KF.  This  member  is  subjected  to  35,200  Ibs.  compression 
and  20,000  Ibs.  tension.  Let  us  assume  2— Ls  5"  x  3i"xTV'  =  5.12n" 
as  section.  Assuming  the  5"  legs  vertical  and  \"  apart  we  have  L/r  = 
177/1.50  =  118,  and  hence  for  the  allowable  compressive  stress  we  have 
16,000-70x118  =  7,740  Ibs.  per  sq.  in.  Then  we  have  35,200 -r  7,740  = 
4.5n/'  for  the  area  required  for  compression,  which  is  0.61n//  less  than 
the  assumed  section. 

For  tension  the  net  area  of  cross-section  of  the  assumed  section  is 
5.12-0.55  =  4.57n".  The  actual  area  required  for  tension  is  equal  to 
20,000 -r  16,000  =  1.25n".  From  the  above  it  is  seen  that  the  assumed 
section  is  somewhat  larger  than  required,  but  owing  to  the  fact  that  it 
is  about  as  satisfactory  as  can  be  obtained  (the  value  of  L/r  being  about 
correct)  it  will  be  used. 

Members  AF  and  FG.  The  member  AF  is  subjected  to  19,300  Ibs. 
tension  while  member  FG  is  subjected  to  22,250  Ibs.  tension  and  13,300 
Ibs.  compression.  The  truss  is  held  transversely  at  G  and  A  by  struts, 
and  as  member  FG  is  in  compression  it  is  necessary  to  consider  (hori- 


DESIGN  OF  BUILDINGS 


615 


zontally)  AG  as  one  member  subjected  to  the  13,300  Ibs.  compression. 
As  the  stresses  are  small  the  designing  is  simply  a  matter  of  obtaining 
the  proper  value  of  L/r.  Let  us  assume  2 — Ls  5"  x  ?>\"  x  yV'  =  5.12n"; 
the  5"  legs  horizontal  and  the  vertical  legs  -J"  apart.  Then  L/r  = 
260/2.44  =  106  in  the  horizontal  plane  and  130/1.03  =  126  in  the  vertical 
plane.  The  first  value  (106)  of  L/r  is  quite  satisfactory,  but  the  last 


•M 


*\1J\J        I'  ^        <H*V<*^   »   VV 

*V-*%W 

^      „  -  24  7$0&*S?  frC 

G//g'-(o  -3£dows  >^jT  ^ 


Fig.   433 


value  is  a  little  high  (125  being  the  limit),  but  as  the  angles  are  fixed 
more  or  less  at  the  joints  in  the  vertical  plane  the  above  assumed  section 
will  be  used.  There  should  be  enough  rivets  in  these  angles  at  points  A 
and  G  to  develop  the  angles  in  tension  in  order  to  obtain  rigidity. 

Members  BF  and  DH.  Each  of  these  members  is  subjected  to  6,710 
Ibs.  compression.  This  stress  is  so  small  that  the  designing  is  simply  a 
matter  of  obtaining  the  proper  value  of  L/r.  Let  us  assume  2 — Ls  2J"  x 
24"xi"  =  2.38n"  for  section.  Then  L/r  =  72/0.78  =  92,  which  is  quite 
satisfactory,  and  hence  the  assumed  section  will  be  used. 

Member  CF.  This  member  is  subjected  to  20,775  Ibs.  tension  and 
24,700  Ibs.  compression.  Let  us  assume  2— Ls  3J"x3i"xTy  =  4.18n" 
as  section  L/r  - 130/1.08  =  120.  Then  for  the  allowable  compressive 
stress  we  have  16,000- 70  x  120  =  7,600  Ibs.  Dividing  this  into  the  com- 
pressive stress  we  obtain  24,700 -f  7,600  =  3.25n"  for  the  area  required 
for  compression.  For  the  net  area  of  cross-section  required  for  tension 
we  have  20,775  -f  16,000  =  1.30n".  From  the  above  it  is  seen  that  the 
assumed  section  is  larger  for  section  than  required,  but  as  the  value  of 
L/r  is  about  as  great  as  is  permissible  the  assumed  section  will  be  used. 


616  STRUCTURAL  ENGINEERING 

Member  CG.  This  member  is  subjected  to  11,800  Ibs.  tension  and 
21,600  Ibs.  compression.  Let  us  assume  2— Ls  4"  x  4"  x  Ty  =  4.80n". 
Then  L/r  =  144/1. 25  =  115,  and  hence  for  the  allowable  compressive  stress 
we  have  16,000-70x115  =  7,950  Ibs.  per  sq.  in.  Dividing  this  into  the 
compressive  stress  we  obtain  21,600  4- 7,950  =  2. 7n//  for  the  area  of  cross- 
section  required  for  compression.  For  the  net  area  of  cross-section 
required  for  tension  we  have  11,800  -f- 16,000  =  0.73n".  From  the  above 
it  is  seen  that  the  assumed  section,  as  far  as  area  is  concerned,  is  larger 
than  required,  but  as  L/r  is  of  about  the  correct  value  the  assumed  section 
will  be  used. 

Member  CH.  This  member  is  subjected  to  5,975  Ibs.  tension.  The 
area  of  cross-section  required  is  about  0.37n".  We  will  use  1 — L  2^"  x 
2J"xJ"  =  2.3S-0.44  =  1.94n"  net  for  section.  This  is  about  as  small  an 
angle  as  can  be  used  and  the  member  be  in  harmony  with  the  other 
members  of  the  truss. 

Members  GH  and  HE.  The  member  GH  is  subjected  to  19,450  Ibs. 
tension  and  10,650  Ibs.  compression,  while  member  HE  is  subjected  to 
25,525  Ibs.  tension  and  9,520  Ibs.  compression.  As  the  truss  is  not  sup- 
ported transversely  at  H  it  is  necessary  to  design  EG  as  one  compression 
member.  Let  us  assume  2— Ls  5"x  3J"x  Ty  =  5.12n"  for  section.  The 
maximum  radius  of  gyration  is  2.44  and  the  least  is  1.03.  Then  L/r  — 
260/2.44  =  106  in  one  case  and  130/1.03  =  126  in  the  other.  For  the 
maximum  allowable  compressive  unit  stress  we  have  16,000  —  70x126  = 
7,180.  Dividing  this  into  the  maximum  compressive  stress  we  obtain 
10,650  -f  7,180  =  1.48n"  for  the  required  area  of  cross-section.  For  the 
required  area  of  cross-section  for  tension  we  have  25,525 -r  16,000  =  1.59n" 
net.  From  the  above  it  is  seen  that  the  assumed  section  is  larger  than  it 
need  be,  as  far  as  section  is  concerned,  but  that  the  value  of  L/r  is  about 
as  large  as  allowed,  so  the  assumed  section  will  be  used. 

Member  GO.  This  member  is  subjected  to  9,900  Ibs.  tension  and 
2,600  Ibs.  compression.  These  stresses  are  so  low  that  the  designing  of 
the  member  is  simply  a  matter  of  obtaining  the  correct  value  for  L/r. 
The  length  of  the  member  is  200  ins.  and  hence  the  radius  of  gyration  of 
the  member  must  not  be  less  than  200/125  =  1.60.  By  using  2 — Ls  5"x 
3J"x&"  and  2— Ls  3J" x 3 J"  x \"  riveted  together  as  shown  at  (a), 
Fig.  433,  we  obtain  about  the  correct  radius  and  hence  this  section  will 
be  used. 

Top  Chord  AB  .  .  .  E.  This  member  should  be  of  the  same  section 
throughout.  It  is  subjected  to  a  maximum  compression  of  43,600  Ibs. 
and  to  a  maximum  tension  of  22,100-8,600  =  13,500  Ibs.  The  purlins 
hold  the  truss  transversely  more  or  less,  but  as  there  is  some  question  as 
to  the  rigidity  thus  obtained  the  top  chord  will  be  considered  unsupported 
transversely  for  half  of  its  length,  which  is  about  18  ft.  or  216  ins.  The 
length  as  regards  the  vertical  direction  is  9  ft.  or  108  ins.  Then,  as 
regards  the  transverse  direction,  the  radius  of  gyration  must  not  be  less 
than  216/125  =  1.73,  and  as  regards  the  vertical  direction  the  radius  of 
gyration  must  not  be  less  than  108/125  =  0.86. 

There  is  a  purlin  concentration  midway  between  panel  points  of 
£  (3,960 +  3,300x0.83)  =  6,700  Ibs.  (see  Figs.  431  and  432)  which  causes 
a  bending  moment  on  the  chord  (considered  as  a  fixed  beam)  of  £  X 
6,700x108  =  90,450  inch  Ibs. 


DESIGN  OF  BUILDINGS  617 

Now,  it  is  seen  that  the  section  of  the  top  chord  must  be  sufficient  to 
transmit  the  direct  stress  and  the  cross  bending  given  above  and  that 
the  radii  must  not  be  less  than  indicated  above. 

Let  us  assume  2— Ls  6"  x4"  x4"  =  9.50n"  for  section;  the  6"  legs 
vertical  and  J"  apart.  The  radii  are  about  1.71  and  1.91,  which  are 
satisfactory  values.  For  the  allowable  compression  we  have  16,000  —  70  x 
216/1.70  =  7,100  Ibs.  per  sq.  in.  The  actual  direct  stress  is  43,600  +  9.5  = 
4,600  Ibs.  per  sq.  in.  and  the  stress  due  to  cross  bending  is  (90,450  x  1.99) 
+  34.8  =  5,180  Ibs.  per  sq.  in.  Now  adding  these  two  stresses  together 
we  obtain  9,780,  which  is  larger  than  the  allowed,  so  the  assumed  section  is 
too  small.  Let  us  try  2— Ls  6"  x  6"  x  f"  =  8.72n".  The  radii  are  about 
2.6  and  1.8.  Then  L/r  in  one  case  is  equal  to  216/2.6  =  83,  and  in  the 
other  108/1.8  =  60.  Then  for  the  allowable  unit  stress  we  have  16,000- 
70x83  =  10,190. 

For  the  stress  due  to  direct  compression  we  have  43,600  +  8.72  = 
5,000  Ibs.'  per  sq.  in.,  and  for  the  stress  due  to  cross  bending  we  have 
(90,450x1.64)4-30.78=4,820.  Now  adding  these  two  stresses  together 
we  have  5,000  +  4,820  =  9,820  Ibs.,  which  is  only  370  Ibs.  less  than  the 
allowable,  so  the  2 — Ls  6"  x  6"  x  f "  will  be  used  as  section  for  the  top 
chord.  We  now  have  all  of  the  truss  members  designed  and  the  sections 
can  be  written  on  the  diagram  as  shown  in  Fig.  433. 

277.  Designing  of  the  Lean-to  Columns. — We  assume  that  these 
columns  are  not  fixed.     The  load  on  each  purlin  due  to  snow  and  dead 
load,  as  given  in  Art.  272,  is  65.5  Ibs.  per  ft.  of  purlin.     The  vertical 
component  of  the  wind  load  is  equal  to  (18  x 4.5)  x  0.908  =  73.5  Ibs.  per 
ft.  of  purlin.     Then  for  the  vertical  load  on  the  rafter  at  each  interme- 
diate purlin  we  have  (65.5  +  73.5)  x  20  =  2,780  Ibs.     Then  the  concentra- 
tion on  the  lean-to  column  due  to  the  roof  load  is  equal  to  2,780  x  3  =  8,340 
Ibs.     Adding  the  weight  of  one-half  of  the  rafter  we  have  8,340+  (13.2  X 
42)  =8,894,  say  8,900  Ibs.,  for  the  total  direct  load  on  the  column. 

The  wind  load  acting  perpendicularly  to  the  column  is  (considering 
it  uniformly  distributed)  equal  to  30x20  =  600  Ibs.  per  ft.  of  column. 
Then  for  the  moment  on  the  column  due  to  this  load  we  have  %  X  600  X  142 
x  12  =  176,400  inch  Ibs.  Let  us  assume  an  8"x34#  H-beam  (Carnegie) 
for  section,  in  which  case  L/r  =168/1.87  =  89.8.  Then  for  the  allowable 
unit  stress  we  have  16,000-70x89.8  =  9.714  Ibs.  and  for  the  actual  direct 
unit  stress  we  have  8,900  +  10  =  890  Ibs.  For  the  maximum  unit  stress 
due  to  cross  bending  we  have 

/=  (176,400  x  4)  + 115.4  =  6,110  Ibs. 

Now  adding  the  direct  unit  stress  to  this  bending  stress  we  have  6,110  + 
890=  7,000  Ibs.  for  the  maximum  unit  stress  on  the  column,  which  shows 
that  the  assumed  H-beam  is  larger  than  required,  but  as  this  beam  seems 
to  be  the  most  satisfactory  section  obtainable  it  will  be  used  throughout 
for  the  lean-to  columns. 

278.  Designing  of  the  Crane  Girders. — The  spacing  and  load  on 

the  wheels  of  a  20-ton  crane  having  a  60-ft.  span  are  given  in  Fig.  414. 
The  maximum  moment  on  the  crane  girder  will  occur  when  the  wheels 
are  in  the  position  shown  at  (o),  Fig.  434.  Taking  moments  about  B 


618 


STRUCTURAL  ENGINEERING 


9-8' 


the  reaction  shown  at  A  is  obtained.     Then  taking  moments  about  wheel 

C  we  have 

28,046x7.58x12  =  2,551,000  inch  Ibs. 

for  the  maximum  moment  on  the  crane  girder  due  to  the  crane.  Assum- 
ing the  crane  girder  to  weigh  80  Ibs.  per  ft.  and  the  rail  J3_0  Ibs.  per  yd., 
making  in  all  100  Ibs.  per  ft.  of  girder,  we  have  J  X  100  X  202  x  12  =  60,000 
inch  Ibs.  for  the  maximum  bending  moment  on  the  girder  due  to  the  weight 

of  the  girder  and  rail.  Adding 
this  moment  to  the  moment  due 
to  the  crane,  we  have  2,551,000  + 
60,000  =  2,611,000  inch  Ibs.  for 
the  total  maximum  bending  mo- 
ment on  the  crane  girder.  Di- 
viding this  by  16,000  we  obtain 
163  for  the  section  modulus 
which  calls  for  a  24"xSO#  I, 
which  will  be  used  for  each 
girder. 

The  maximum  reaction  on  the 
crane  girder  will  occur  when  the 
crane  wheels  are  in  the  position 
shown  at  (6),  Fig.  434.  Tak- 
ing moments  about  E  we  obtain 
the  reaction  56,100  Ibs.  shown 
at  D,  which  is  the  maximum  end 
shear  on  the  crane  girder  due 
to  the  crane  wheels.  The  end 
shear  due  to  the  weight  of  the 
crane  girder  and  rail  is  equal  to 
100  +  20/2  =  1,000  Ibs.  Then 


AI 

1 

Ki 

(P)                       O 

7.56' 

i 

2.42 

2-42. 

7.58' 

4     C3- 
20' 

(a) 


r  *<-s>' 

/o-  4-" 

\                               20  '-  0  " 

(b) 

Fig.  434 


for  the  total  end  shear  or  reaction  on  the  crane  girder  we  have  56,100  + 
1,000  =  57,100  Ibs. 

279.  Designing  of  the  Main  Columns.— The  load  applied  to  the 
top  of  each  intermediate  column  due  to  snow  and  dead  load  is  13,200  Ibs., 
as  given  in  Fig.  431.  The  load  from  the  lean-to  rafter  due  to  snow  and 
dead  load  is  equal  to  65.5  x  20  x  3  =  3,930,  say  4,000  Ibs.  The  direct  stress 
due  to  the  wind  load  is  equal  to  the  vertical  component  of  the  reaction  at 
the  point  of  contra-flexure,  the  maximum  of  which  is  given  in  Fig.  432  as 
11,500  Ibs.  Then  for  maximum  direct  stress  on  the  column  above  the 
crane  girder  we  have  13,200  +  4,000  +  11,500  =  28,700  Ibs.  This  load 
comes  on  the  leeward  column.  As  is  seen  from  Fig.  432,  the  maximum 
moment  on  the  column  due  to  wind  load  occurs  at  the  knee  brace  and  is 
equal  to  (//2  x  12.5)  12  =  11,500  x  12.5  x  12  =  1,725,000  inch  Ibs.  Let  us 
assume  the  part  of  the  column  above  the  crane  girders  to  be  a  built 
I-section  composed  of  4— Ls  6"  x  3 \"  x  TV  and  a  web  18"  xf".  The 
distance  from  the  truss  down  to  the  crane  girder,  which  is  16  ft.  or  192 
ins.  (see  Fig.  429),  will  be  taken  as  the  length  of  this  column.  The  least 
radius  of  gyration  of  the  section  is  2.5.  Then  we  have  L/r  =  192/2.5  = 
76.8,  and  hence  for  the  allowable  unit  stress  we  have  16,000-  (70  x  76.8) 
=  10,624  Ibs.  The  moment  of  inertia  of  the  section  with  reference  to  the 


DESIGN  OF  BUILDINGS 


619 


H 


axis  perpendicular  to  the  web  is   1,576.     Then    for  the  maximum  unit 
stress  due  to  cross  bending  we  have 

/  =  (1,725,000  x  9)  + 1,576  =  9,850  Ibs. 

The  area  of  the  cross-section  is  26.87n".  For  the  direct  stress  on  the 
column  we  have  28,700  +  26.87  =  1,068  Ibs.  per  sq.  in.  Adding  this  to  the 
bending  stress  we  obtain  9,850  +  1,068  =  10,918  Ibs.  for  the  total  maximum 
unit  stress  on  the  column  (above  crane 
girder),  which  is  only  294  Ibs.  greater 
than  the  10,624  Ibs.  allowable,  so  the 
above  assumed  section  will  be  used. 

The  part  of  the  column  below  the 
crane  girder  will  be  at  least  11  ins.  wider 
than  the  part  above  (see  Fig.  414),  so 
that  the  crane  girder  can  be  connected 
directly  to  the  column.  This  requires 
that  the  part  below  the  crane  girder  be 
30"  wide.  So  let  us  assume  the  part  of 
the  column  below  the  crane  girder  to 
be  composed  of  4— Ls  6"  x  3£"  x  TV  = 
20.12n"  and  a  web  30"  x  A"  =  9.38, 
making  in  all  29. 5n"  of  cross-section. 
The  least  radius  of  gyration  is  2.55. 
The  distance  from  the  crane  girder  to 
the  base  of  the  column  is  19'-0"  or  228". 
So  we  have  L/r  =  228/2.55  =  89.4,  and 
hence  for  the  allowable  compression  we 
have  16,000-70x89.4  =  9,742  Ibs.  per 
sq.  in.  The  direct  load,  considering  the 
windward  column,  applied  from  the  part 
of  the  column  above  the  crane  girder,  is 
equal  to  13,200  +  4,000  +  1,700  =  18,900 
Ibs.  This  causes  a  direct  unit  stress  of 
18,900  +  29.5  =  641  Ibs.  The  two  angles 
of  the  column  directly  under  the  crane 
girders  should  be  sufficient  to  carry  the 
maximum  concentration  from  these  gird- 
ers. As  the  crane  wheels  are  equal  in 
weight  the  maximum  concentration  due 
to  the  wheels  is  equal  to  the  maximum 
end  shear  on  the  crane  girder,  which  is 
given  in  Art.  272  as  56,100  Ibs.  The 
concentration  due  to  the  weight  of  the 
crane  girders  and  struts  is  equal  to 
(80  +  20)  20  =  2,000  Ibs.  Then  for  the 
total  concentration  on  the  column  from 
the  crane  girders  we  have  56,100  + 
2,000  =  58,100  Ibs.  The  area  of  cross- 
section  of  the  two  angles  is  10.06n". 
Then  we  have  58,100  +  10.06  =  5,776  Ibs.  for  the  direct  stress  on  the  two 
angles  due  to  the  concentration  from  the  crane  girders.  Adding  to  this 


(a) 


Fig.   435 


620  STRUCTUKAL  ENGINEERING 

the  direct  stress  transmitted  from  the  part  of  the  column  above  the  crane 
girder  we  obtain  5,776  +  641  =  6,417  Ibs.  for  the  maximum  direct  stress 
on  the  two  column  angles  directly  under  the  crane  girders. 

As  is  seen  from  Fig.  4:32,  the  maximum  bending  moment  on  the  part 
of  the  column  below  the  crane  girder  due  to  wind  is  equal  to  12.5  X 
11,500x12  =  1,725,000  inch  Ibs.,  which  occurs  at  the  base  of  the  column. 
The  moment  of  inertia  of  the  column  in  reference  to  the  gravity  axis 
perpendicular  to  the  web  is  about  4,800.  Then  we  have 

/  =  (  1,725,000  x  15)  -r  4,800  =  5,390  Ibs.  per  sq.  in. 

for  the  maximum  stress  due  to  cross  bending.  Adding  this  to  the  above 
direct  stress  we  have  6,417  +  5,390  =  11,807  Ibs.  for  the  total  maximum 
unit  stress  on  the  two  angles  directly  under  the  crane  girders.  The 
allowable  stress  is  given  above  as  9,742  Ibs.  per  sq.  in.  ;  but  as  the  maxi- 
mum crane  load  and  maximum  wind  load  are  not  likely  to  occur  at  the 
same  time  this  stress  could  safely  be  increased  25  per  cent.  So  we  have 
1.25x9,742  =  12,179  Ibs.  for  the  unit  stress  permissible  for  the  combined 
loading.  This  shows  that  the  assumed  section  for  the  part  of  the  column 
below  the  crane  girder  is  about  correct  and  hence  will  be  used.  The  two 
outside  angles  and  web  of  the  column  are  stressed  less  than  the  angles 
directly  under  the  crane  girders,  but  it  is  desirable  to  have  a  symmetrical 
section  and  hence  the  main  section  of  the  column  as  given  above  is  satis- 
factory throughout. 

280.  Designing  of  Column  Base.  —  Case  I.  When  column  is  sub- 
jected to  direct  load  only.  In  that  case  the  pressure  on  the  base  will  be 
uniformly  distributed  as  indicated  at  (a),  Fig.  435. 

Let  P  =  direct  load  on  column, 

p  =  pressure  per  square  in., 

6  =  width  of  base  in  ins., 

d  —  length  of  base  in  ins. 
Then  we  have 


CD 


The  value  of  p  should  not  exceed  600  Ibs.,  which  is  the  allowable 
pressure  on  concrete  masonry. 

Case  II.  When  the  column  is  subjected  to  both  direct  load  and  cross 
bending  but  the  bending  not  sufficient  to  reverse  the  direct  pressure.  In 
that  case  the  pressure  will  vary  as  indicated  at  (6),  Fig.  435. 

LetP=  direct  load  on  column, 

p=  maximum  pressure  per  sq.  in., 
p'  —  minimum  pressure  per  sq.  in., 
M—  bending  moment  in  in.  Ibs., 
6  =  width  of  base  in  ins., 
d—  length  of  base  in  ins. 

Then  we  have 


DESIGN  OF  BUILDINGS  621 

=  P      6M 
and 


,  ,  ,  -j, (2). 

bd      bd~ 


Case  III.  When  the  column  is  subjected  to  both  direct  load  and 
cross  bending  and  the  bending  reverses  the  direct  pressure.  In  that  case, 
assuming  linear  variation,  the  forces  will  be  as  indicated  at  (c),  Fig.  435. 

Let  P  =  direct  load  on  column, 

p  —  maximum  pressure  per  sq.  in., 

/  =  stress  per  sq.  in.  on  the  anchor  bolt  =10,000  Ibs., 

A  =  area  of  cross-section  of  anchor  bolt  or  bolts  on  each  side  of 

column, 

M  =  bending  moment  in  in.  Ibs., 

n  —  ratio  of  the  modulus  of  elasticity  of  steel  to  masonry  =  15, 
b  =  width  of  column  base  in  ins., 
d  —  length  of  column  base  in  ins. 

The  problem  is  to  determine  p  and  A.  As  is  seen  from  the  diagram 
at  (c)  these  could  readily  be  determined  if  x  were  known.  So  we  will 
first  derive  an  expression  for  the  value  of  x. 

It  is  obvious  that  the  moments  of  the  forces  about  o  (center  of  the 
column)  must  be  equal  to  M. 
Then  we  have 


As  the  summation  of  the  vertical  forces  must  equal  0,  we  have 

fx6-/*=P  ...................................  .  .....  (5) 

From  the  stress  diagram  at  (c)  we  have 

P  x 


'  d-x' 

v»; 

from  which  we  obtain 


From  (5)  we  obtain 


From  (4),  (5),  and  (6)  the  following  expression  for  x  can  be  obtained: 

=  0  ............  (8) 


jo  of 


622  STEUCTURAL  ENGINEERING 

The  value  of  x  for  any  case  can  be  determined  from  (8)  and  then 
the  value  of  p  can  be  obtained  from  (6)  and  A  from  (7).  For  conven- 
ience the  anchor  bolts  are  assumed  to  be  exactly  at  the  edge  of  the  base 
plate,  which  of  course  is  not  absolutely  true,  but  the  error  resulting  from 
the  assumption  is  small. 

Now,  considering  the  windward  column,  the  direct  load  above  the 
crane  girder  as  previously  given  is  28,700  Ibs.  and  the  load  from  the 
crane  girder  is  58,100  Ibs.,  and  assuming  the  column  to  weigh  2,000  Ibs. 
we  have 

P  =  28,700  +  58,100  +  2,000  =  88,800  Ibs. 

for  the  direct  load  on  the  column.  As  previously  found  the  bending 
moment 

M  =  1,725,000  inch  Ibs. 

Let  us  assume  d  equals  40"  and  b  equals  16".  Substituting  these  values  in 
(8)  and  assuming  /  equals  10,000  we  obtain 

x  —  20  ins.  (about). 
Substituting  in  (6)  we  obtain 

p  =  6661bs., 

which  is  quite  satisfactory,  as  the  pressure  is  due  to  direct  load  and  cross 
bending  which  are  not  likely  to  occur  simultaneously.  In  fact  p  could 
safely  be  800  or  900  Ibs. 

Substituting  in  (7)  we  obtain 

A  =  (^  x  20  x  16  -  88,800  \  * 10,000  =  2.77  sq.  ins. 

for  the  area  of  cross-section  of  anchor  bolts.  Use  two  If"  anchor  bolts 
on  a  side. 

281.  Drawings. — The  data  given  in  Articles  270  to  280  is  suffi- 
cient for  making  the  stress  sheet,  Fig.  436,  except  for  the  gable  fram- 
ing, girts,  and  bracing,  which  can  be  considered  more  conveniently 
as  the  work  on  the  stress  sheet  progresses.  The  gable  framing  is  de- 
signed to  resist  a  wind  pressure  of  30  Ibs.  per  sq.  ft.  on  the  end  of  the 
building.  Each  gable  column  can  be  considered  as  a  vertical  beam  35  ft. 
long  and,  as  they  are  about  11  ft.  apart,  the  load  on  each  is  11x30  =  330 
Ibs.  per  ft.  of  length.  Then  the  bending  moment  on  each  is  £x330x 
352x  12  =  606,370"*.  Dividing  this  by  16,000  we  obtain  37.9  for  the 
section  modulus.  This  calls  for  a  12"x35#  I  which  will  be  used  for 
each  gable  column. 

The  bracing  between  the  trusses  and  main  columns  in  the  end  bays 
should  be  sufficient  to  resist  the  wind  pressure  on  the  end  of  the  building. 
The  stresses  in  this  bracing  are  determined  as  indicated  in  Fig.  437.  The 
diagram  at  (a)  represents  the  bracing  between  the  main  columns  and  the 
diagram  at  (6)  represents  the  stresses  in  the  same.  The  diagram  at  (c) 
represents  the  bracing  in  the  plane  of  the  roof  and  the  diagram  at  (d) 
represents  the  bracing  in  the  plane  of  the  bottom  chord  of  trusses.  The 
designing  of  the  bracing  in  intermediate  bays  is  mostly  a  matter  of  judg- 
ment as  it  is  intended  mostly  for  rigidity. 


It'! 
f-ftH 


rrM 


623 


624 


STRUCTURAL  ENGINEERING 


The  girts  are  designed  to  resist  a  horizontal  wind  pressure  of  30  Ibs. 
per  sq.  ft.  and  in  some  cases  they  should  act  as  struts,  that  is,  L/r  should 
not  exceed  125.  After  the  stress  sheet,  Fig.  436,  is  completed  the  shop 


20-0' 


(a) 


drawings  and  bills  for  the  structure  can  be  made.     A  fair  idea  of  the 
details  can  be  obtained  from  the  general  drawing,  Fig.  438. 

HIGH  BUILDINGS 

282.  Preliminary. — In  the  case  of  modern  high  buildings  such  as 
office  buildings,  hotels,  etc.,  all  loads  including  exterior  walls,  partitions, 
floors,  and  live  loads  are  practically  in  every  case  supported  upon  a  steel 
frame  which  consists  of  columns  and  beams.  While  the  designing  of 


626  STRUCTURAL  ENGINEERING 

these  buildings  as  a  whole  is  architectural  work  the  designing  of  the  steel 
frame  is  purely  a  structural  engineering  problem  and  hence  will  be  herein 
treated  as  such. 

283.  Dead  Load. —  The  dead  load  includes  all  permanent  parts  of 
the  building  and  also  all  permanent  fixtures  and  mechanisms  supported 
by  the  building.      In  estimating  dead  loads  the  following  units  of  weights 
may  be  used: 

Steel 490  Ibs.  per  cu.  ft. 

Concrete 150  Ibs.  per  cu.  ft. 

Brick  walls   140  Ibs.  per  cu.  ft. 

Hollow  tile  walls 40  Ibs.  per  cu.  ft. 

Hollow  tile  floors   (tile  alone)..  50  Ibs.  per  cu.  ft. 

Wood    4.5  Ibs.  per  sq.  ft.  B.  M. 

Plaster    5  Ibs.  per  sq.  ft. 

The  weights  of  the  other  materials  can  be  obtained  from  handbooks. 

284.  Floor  Load. — The  floor  load  consists  of  the  weight  of  the 
floor  proper  and  the  load  supported  on  the  floor.     The  former  is  usually 
referred  to  as  the  dead  load  and  the  latter  as  live  load.     The  dead  load 
is  always  first  assumed  and  then  verified  by  calculations.     The  live  load, 
as  a  rule,  can  only  be  estimated.     The  building  ordinances  of  the  differ- 
ent cities  specify  what  these  loads  shall  be  in  each  case.      In  the  case  of 
office  buildings  the  live  load  for  the  first  floor  is  usually  taken  as  100  to 
150  Ibs.  per  sq.  ft.  of  floor;  the  other  floors  from  40  to  75  Ibs.  per  sq.  ft. 

Designing  of  a  Ten-Story  Office  Building 

285.  Data:— 

Length  =  7   bays    @    15'-0"      =  105'-0" 
Width    =4  bays    @    15'-0"      =    60'-0" 

f  Basement  =    lO'-O"  1 

Height  =1   1st  story  =    18'-0"  Il36'-0" 

1  9  stories  @  12'-0"  =  108'-0"  J 
Dead  load  to  be  computed. 

f  First  floor — 100  Ibs.  per  sq.  ft.  of  floor. 
Live  load  •<  All  other  floors — 50  Ibs.  per  sq.  ft.  of  floor. 

I  Roof — 25  Ibs.  per  sq.  ft.  of  roof. 

Wind  load — 20  Ibs.  per  sq.  ft.  of  vertical  exposed  surface. 
Type  of  floor — solid  concrete. 
All  metal  to  be  incased  in  concrete. 

In  order  to  resist  the  wind  pressure  on  the  building  the  floor 
beams  will  be  rigidly  connected  to  the  columns.  These  beams  will 
be  considered  as  fixed  beams  for  all  loads. 

The  floor  framing  in  general  will  be  as  shown  in  Fig.  439. 
Allowable  unit  stresses  will  be  the  same  as  for  highway  bridges 
(see  12  of  Art.  222). 

286.  Wind  Stresses.— As  seen  from  Fig.  439,  it  is  necessary  to 
compute  the  wind  stresses  in  bents  BE,  CC,  and  EE,  at  least.     The  wind 
stresses  in  bents  AA  will  be  one-half  as  much  as  in  bents  BE,  and  like- 
wise the  stresses  in  bent  DD  will  be  half  as  much  as  in  bents  EE  and  FF. 


DESIGN  OF  BUILDINGS 


627 


Let  us  first  consider  bents  BB,  the  elevation  of  which  is  shown  in 
Fig.  440.     The  wind  load  at  the  roof  is  equal  to 

20  x  15  x  (6 +  4)  =3,000  Ibs. 

At  each  floor  from  the  tenth  down  to  the  third  (inclusive)  the  wind  load 
is  equal  to  20x15x12  =  3,600  Ibs.  as  shown,  and  at  the  second  floor  it  is 

60-0" 


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Fig.  439 

equal  to  20xl5x  (6 +  9)  =4,500  Ibs.,  and  at  the  first  floor  it  is  equal  to 
20x15x9  =  2,700  Ibs.  Having  the  wind  loads  determined  we  can  pro- 
ceed with  the  determination  of  the  stresses  due  to  same  by  beginning  at 
the  roof  and  working  downward  story  after  story. 

The  columns  and  the  floor  beams  connecting  to  them  will  be  con- 
sidered fixed  and  the  point  of  contra-flexure  in  each  column  of  each  story 
will  be  considered  to  be  at  mid-story,  while  the  point  of  contra-flexure  in 
each  floor  beam  will  be  considered  to  be  at  mid-span.  Any  part  of  the 
structure  between  points  of  contra-flexure  can  be  considered  as  an  inde- 
pendent structure. 

In  determining  the  stresses  the  building  as  a  whole  will  be  considered 


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Fig.  440 

628 


DESIGN  OF  BUILDINGS 


629 


to  act  as  a  vertical  cantilever  beam,  in  which  case  the  direct  stress  on 
the  columns  will  vary  directly  as  their  distance  from  the  neutral  axis. 

Considering  the  part  of  the  structure  above  the  points  of  contra- 
flexure  of  the  columns  in  the  tenth  story  we  have  the  independent  struc- 
ture shown  at  (a),  Fig.  441.  As  is  obvious,  the  neutral  axis  will  be  at 


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Fig.   441 

Ox,  the  center  of  the  building.     Then  the  vertical  reaction  on  the  columns 
will  vary  directly  as  their  distance  out  from  0±  as  indicated  at  (6). 
For  the  moment  of  the  wind  load  about  Ox  we  have 

M  =  3,000x6-18,000  ft.  Ibs. 

This  moment  must  be  balanced  by  the  moment  of  the  vertical  reactions  on 
the  columns.  Let  R  represent  the  reaction  at  F.  Then  the  reaction  at 
a  point  one  foot  (unit)  out  from  Ol  would  be  12/30,  and  hence  the  reac- 
tion at  G  and  H  will  be  (12/30)  15  and  similarly  at  F  and  K  it  will  be 
(12/30)  30.  Then  taking  moments  about  0±  we  have 


from  which  we  obtain 


<  5 


The  number  75  is  a  constant  that  can  be  used  in  determining  the  12  In  all 
the  other  stories. 


630  STRUCTURAL  ENGINEERING 

Now  having  the  vertical  reaction  R  at  F  determined,  we  can  readily 
determine  the  reactions  on  the  other  columns  as  the  intensity  in  each  case 
is  directly  proportionate  to  the  distance  of  the  column  from  0^  Then, 
for  the  reaction  at  G  and  H,  we  have  (15/30)  x  240  =  120  Ibs.  and  for 
the  reaction  at  K  we  have  (30/30)  x  240  =  240  Ibs.,  which  of  course  is 
the  same  as  at  F. 

Having  the  vertical  reactions  on  the  columns  determined  we  can  next 
determine  the  vertical  shear  on  the  floor  beams  by  simply  adding  up  the 
vertical  forces  beginning  at  either  A  or  E.  For  example,  beginning  at  A 
we  have 

Fl  =  240 

for  the  shear  on  girder  AB.     For  the  shear  on  girder  EC  we  have 

F2  =  240 +  120  =  360. 

The  structure  being  symmetrical  about  Oi  the  shears  and  reactions  on 
one  side  of  Oj  will  be  equal  and  opposite  to  those  on  the  other  side,  and 
hence  only  one-half  of  the  structure  need  be  considered  in  determining 
these  stresses. 

Having  the  vertical  reactions  on  the  columns  and  the  vertical  shear 
on  the  floor  beams  determined,  the  horizontal  shears  H\,  H2,  and  H3  on 
the  columns  can  be  computed.  Considering  the  part  of  the  structure 
shown  at  (c)  as  an  independent  structure  and  taking  moments  about  A 
we  have 

240x7.5-7/1x6  =  0, 
from  which  we  obtain 

#1  =  300  Ibs. 

Then  summing  up  the  horizontal  forces,  we  have 

3,000-  300  -F  =  0, 

from  which  we  obtain 

F  =  2,700  Ibs. 

for  the  direct  compression  in  beam  AB.  Considering  the  part  of  the 
structure  shown  at  (d)  as  an  independent  structure  and  taking  moments 
about  JB,  we  have 

240  x  7.5  f  360  x  7.5  -  H2  x  6  =6, 

from  which  we  obtain 

#2  =  750  Ibs. 

Then  summing  up  the  horizontal  forces  we  have 

2,700- 750 -Fl  =  0, 
from  which  we  obtain 

Fl  =  1,950  Ibs. 

for  the  direct  compression  in  floor  beam  BC.  Likewise,  considering  the 
part  of  the  structure  shown  at  (e)  as  an  independent  structure  and  taking 
moments  about  C  we  have 

360x15-^3x6  =  0, 


DESIGN  OF  BUILDINGS 


631 


from  which  we  obtain 

H3  =  900  Ibs. 

Then  adding  up  the  horizontal  forces  we  have 

1,950-  900  -F2  =  0, 
from  which  we  obtain 

F2  =  1,050  Ibs. 

for  the  direct  stress  in  beam  CD.  The  direct  stress  in  beam  DE  is  equal 
to  1,050-750  =  300  Ibs.,  which  is  and  should  be  equal  to  HI. 

These  shears  and  stresses  can  now  be  written  on  the  diagram  in 
Fig.  440. 

Next  considering  the  part  of  the  structure  between  the  points  of 
contra-flexure  in  the  columns  of  the  tenth  and  ninth  stories  we  obtain  the 
independent  structure  shown  at  (a),  Fig.  442.  Taking  moments  about 


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Fig.  442 


O2  of  the  wind  forces  on  the  building  above  this  point  (see  Fig.  440),  we 
have 

M  =  3,000  x  18  +  3,600  x  6  =  75,600  ft.  Ibs. 

Letting  R  represent  the  vertical  reaction  on  the  column  at  F  we  have 


from  which  we  obtain 


632  STRUCTURAL  ENGINEERING 

Then  the  reaction  on  the  column  at  G  is  equal  to  1,008  (15/30)  =504  Ibs. 
Beginning  at  A  and  adding  up  the  vertical  forces  (algebraically) 
we  have  1,008-240  =  768  Ibs.  for  the  shear  on  beam  AB,  1,008-240  + 
504-120  =  1,152  Ibs.  for  the  shear  on  beam  EC.  Now  considering  the 
part  of  the  structure  shown  at  (c)  as  an  independent  structure  and  taking 
moments  about  A,  we  have 

HI  x  6  +  300  x  6  -  768  x  7.5  =  0, 

from  which  we  obtain 

HI  =  660  Ibs. 

for  the  horizontal  shear  on  the  column  at  F. 

Similarly,  considering  the  part  of  the  structure  shown  at  (d)  as  an 
independent  structure  and  taking  moments  about  B,  we  have 

H2  x  6  +  750  x  6  -  768  x  7.5  -  1,152  x  7.5  =  0, 

from  which  we  obtain 

H2  =  1,650  Ibs. 

Considering  the  part  of  the  structure  shown  at  (b)  as  an  independent 
structure  and  taking  moments  about  C,  we  have 

H3  x  6  +  900  x  6  - 1,152  x  15  =  0, 

from  which  we  obtain 

H  3  =  1,980  Ibs. 

for  the  shear  on  the  column  at  02.  Now  adding  up  the  horizontal  forces 
beginning  at  A  we  have  3,600  +  300  -  660  =  3,240  Ibs.  for  the  direct  com- 
pression in  beam  AB,  3,600  +  300-660-1,650  +  750  =  2,340  Ibs.  for  the 
direct  compression  in  beam  BC,  2,340  +  900-1,980  =  1,260  Ibs.  for  the 
direct  compression  in  beam  CD,  and  1,260  +  750-1,650  =  360  Ibs.  for 
the  direct  compression  in  beam  DE. 

Continuing  in  the  manner  shown  above,  considering  the  parts  of  the 
structure  between  points  of  contra-flexure  in  the  column  of  consecutive 
stories  on  down  to  the  base  of  the  building,  the  direct  stresses  and  shears 
on  the  columns  and  floor  beams,  as  shown  in  Fig.  440,  can  be  determined. 

It  will  be  seen  from  Fig.  440  that  the  horizontal  shear  on  each 
column  varies  from  the  ninth  down  to  the  second  story  by  a  constant  and 
chat  the  vertical  shear  on  the  floor  beams  in  each  bay  varies  from  the 
tenth  down  to  the  third  floor  by  a  constant,  and  hence  these  stresses  can 
be  quickly  computed  by  the  use  of  the  constants,  which  can  be  determined 
after  the  stresses  are  computed  in  the  three  top  stories.  Also,  it  will  be 
seen  (see  Fig.  440)  that  the  direct  stress  in  the  floor  beams  in  each  bay 
is  the  same  from  the  tenth  down  to  the  third  floor.  After  the  shears  on 
the  columns  and  floor  beams  are  computed  by  the  use  of  the  constants, 
the  direct  stress  on  the  columns  can  be  determined  by  beginning  at  the 
ninth  story  and  adding  up  the  vertical  forces  included  between  the  points 
of  contra-flexure.  As  an  example,  for  the  direct  stress  in  the  outer  left- 
hand  column  of  the  sixth  story,  we  have  4,272  +  2,496  =  6,768  Ibs.  For 
the  next  column  to  the  right  we  have— 2,496  +  2,136  +  3,744  =  3,384  Ibs., 
and  so  on.  To  obtain  the  stresses  in  the  second  story  floor  beams  and 


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633 


634 


STRUCTURAL  ENGINEERING 


first  story  columns  we  first  take  moments  about  O10  of  the  wind  forces 
above  that  point.  For  this  moment  (see  Fig.  440),  we  have 

M  =  3,000  x  117  +  28,800  x  63  +  4,500  x  9  =  2,205,900  ft.  Ibs. 

Then  dividing  this  by  the  constant  75  (previously  determined)  we  obtain 
29,412  Ibs.  for  the  reaction  or  direct  stress  in  the  outer  left-hand  column. 
The  reactions  on  the  other  columns  can  then  be  quickly  determined  by 
direct  proportion,  and  the  remainder  of  the  work  of  determining  the  stress 
is  the  same  as  previously  explained. 

To  obtain  the  stresses  in  the  floor  beams  of  the  first  floor  and  in  the 
basement  columns  we  would  first  take  moments  about  0^  (see  Fig.  440) 
of  the  wind  forces  above  this  point.  For  this  moment,  we  have 

M  =  3,000  x  131  +  28,800  x  77  +  4,500  x  23  +  2,700  x  5  -  2,727,600  ft.  Ibs. 

Then  dividing  this  by  the  constant  75  we  obtain  36,368  Ibs.  for  the  reac- 
tion or  direct  stress  on  the  outer  left-hand  column.     The  reactions  on  the 
other  columns  are  readily  determined  by  direct  proportion,  and  the  re- 
mainder of  the  work  of  de- 

he  termining    the    stresses    is 

the  same  as  previously  ex- 
plained. 

From  the  above  it  is 
seen  that  all  of  the  direct 
stresses  and  shears  in  the 
columns  and  floor  beams 
of  the  bent  are  obtained  by 
fully  analyzing  only  the 
three  top  stories  and  the 
first  floor  and  basement. 

We  will  next  consider 
the  bents  CC  (through  the 
light  court),  the  elevation 
of  which  is  shown  in  Fig. 
443.  Considering  the  part 
of  the  structure  above  the 
points  of  contra-flexure  of 
the  columns  in  the  tenth 
story  we  obtain  the  inde- 
pendent structure  which  is 
shown  at  (a),  Fig.  444. 
As  these  bents  are  unsym- 
metrical  we  have  to  deter- 
mine the  location  of  the 
neutral  axis  00.  The  re- 
actions on  the  columns  will  vary  directly  as  their  distance  from  this 
neutral  axis,  as  indicated  at  (6),  and  the  sum  of  the  reactions  on  one 
side  must  equal  the  sum  of  those  on  the  other. 

Let  2  =  the  distance  from  A  to  the  neutral  axis,  and  for  the  sake  of 
simplicity  let  a,  b,  and  c  represent  the  bay  lengths  as  indicated,  and  let  R 
represent  the  reaction  at  E. 


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Fig.   444 


DESIGN  OF  BUILDINGS  635 

Then  we  have  R  for  the  reaction  at  E,  (R/z)  (2  -  a)  for  the  reaction 
at  G,  (R/z)  (a  +  b-z)  for  the  reaction  at  H,  and  (R/z)  (a  +  b  +  c-z) 
for  the  reaction  at  K.  Then,  as  the  sum  of  the  reactions  on  one  side  of 
the  neutral  axis  must  be  equal  to  the  sum  of  those  on  the  other  side  of  the 
neutral  axis,  we  have 


from  which  we  obtain 


for   the   distance    from  A    to   the   neutral    axis.     Now,   substituting   the 
numerical  values  of  the  bay  lengths  in  (1)  we  obtain 


which  gives  the  location  of  the  neutral  axis. 
Taking  moments  about  Oi  we  have 


3,000x6  =  0, 
from  which  we  obtain 


for  the  reaction  on  the  column  at  E.  The  number  43.23  is  the  constant 
for  determining  the  reaction  R  for  all  the  other  stories.  Having  the 
reaction  at  E  determined  the  reaction  on  the  other  columns  is  readily 
determined  by  direct  proportion.  For  example,  the  reaction  at  H  is 
(416/21.25)  8.75  =  171  Ibs.  After  the  reactions  on  the  columns  are  deter- 
mined the  shears  on  the  columns  and  the  shears  and  direct  stresses  on  the 
floor  beams  can  be  determined,  as  previously  explained,  by  considering 
the  independent  portions  shown  at  (c),  (d),  (e),  and  (/).  In  fact,  the 
remainder  of  the  work  of  determining  the  direct  stresses  and  shears  on 
the  columns  and  floor  beams  shown  in  Fig.  443  is  the  same  as  previously 
explained  for  bents  BB. 

The  shears  and  direct  stresses  in  the  columns  and  floor  beams  for 
bents  EE  and  FF  are  given  in  Fig.  445.  These  are  determined  in  the 
same  manner  as  previously  shown  for  bents  BB.  It  will  be  seen  that  the 
neutral  axis  for  bents  EE  and  FF  is  midway  between  the  central  columns. 
After  the  shears  and  direct  stresses  on  the  columns  and  floor  beams,  as 
given  in  Figs.  440,  443,  and  445,  are  determined  the  stresses  at  the  column 
connections  are  readily  obtained,  as  will  be  seen  later. 

287.  Designing  of  the  Roof  Framing.  —  The  columns  and  beams 
will  be  spaced  as  shown  in  Fig.  446. 

The  roof  covering  will  be  a  4"  solid  concrete  slab. 

The  assumed  loads  on  the  roof  will  be  as  follows: 


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Fig.  445 

636 


DESIGN  OF  BUILDINGS 


637 


Live  load  (Chicago)  ............  25  Ibs.  per  sq.  ft.  of  roof. 


Dead  loadTConcrete   slab  .......    50  lbs*  per  sq<  ft>  of  roof- 

\Suspended  ceiling.  ...    1^0  Ibs.  per  sq.  ft.  of  roof. 

60  Ibs.  per  sq.  ft.  of  roof. 


C2 


C3 


C3 


C3 


C2 


r-  7  "x  /s  < 

/ 


Cl 


'cj- 


c/ 


C/     2          .  .C/ 


C4 


C4- 


C/ 


Roof 

Fig.   44G 

Beams  1.     The  dead  load  on  each  of  these  beams  is  as  follows 

Concrete  slab  and   ceiling  =  60x5  =  300  Ibs.  per  ft.  of  beam. 
Concrete  around  beam  75  Ibs.  per  ft.  of  beam. 

Beam    (assumed)  =  15  Ibs.  per  ft.  of  beam. 

390  Ibs.  per  ft.  of  beam. 
Then,  for  the  maximum  bending  moment  due  to  dead  load  we  have 


(538  STRUCTURAL  ENGINEERING 

J  x  390  x  1?  x  12  =  131,625  inch  Ibs. 

The  live  load  is  25  x  5  =  125  Ibs.  per  ft.  of  beam.  Then,  for  the  live  load 
moment  on  each  beam  we  have 

•Jxl25xl52x  12=42,187  inch  Ibs. 

Adding  the  two  moments  together  we  obtain  131,625  +  42,187  =  173,812 
inch  Ibs.  for  the  total  maximum  moment  on  the  beam.  Dividing  this  by 
16,000  we  obtain  10.8  for  the  section  modulus  which  calls  for  a  7"xl5# 
I-beam,  which  will  be  used. 

T-,    ,    ,        /Dead  load  =  390x7.5  =  2,925  Ibs. 
End  shear 


3,862  Ibs. 

Beams  2.     These  beams  will  be  connected  rigidly  to  the  columns  to 
resist  wind  pressure  and  consequently  will  be  fixed  beams  for  all  loads. 
The  dead  load  on  each  of  these  beams  is  as  follows: 

Fire  wall  [see  sketch  at  (6)  ]  =  1  x  140  x  4  =  560  Ibs.  per  ft.  of  beam. 
Concrete  slab  and  ceiling  =  60  x  2.5  150  Ibs.  per  ft.  of  beam. 
Concrete  around  beam  =  120  Ibs.  per  ft.  of  beam. 

Beam  (2  —  [s)  =  30  Ibs.  per  ft.  of  beam. 

860  Ibs.  per  ft.  of  beam. 

Then  for  the  maximum  bending  moment  at  the  ends  of  the  beam  due 
to  dead  load,  considering  each  bracket  to  be  18  ins.  long,  we  have 
A  x  860  x!22x  12  =  123,840  inch  Ibs.,  and  one-half  of  this,  or  61,920  inch 
Ibs.,  will  be  the  moment  at  the  center  of  span  (see  Art.  69).  The  live 
load  on  the  beam  is  2.5x25  =  62.5  Ibs.  per  ft.  of  beam.  Then  for  the 
bending  moment  at  the  ends  of  the  beam  due  to  live  load  we  have  y1^  x 
62.5  x!2"2x  12  =  9,000  inch  Ibs.,  and  one-half  of  this,  or  4,500  inch  Ibs., 
will  be  the  moment  at  the  center  of  span. 

The  maximum  shear  on  these  girders  due  to  wind  pressure  is  one-half 
of  the  maximum  given  in  Fig.  440,  or  360  +  2  =  180  Ibs.  Then  for  the 
moment  at  the  end  of  the  beam  (at  end  of  bracket)  we  have  180  x  72  = 
12,960  inch  Ibs. 

Now  combining  the  maximum  dead  and  live  moments  we  have 
123,840  +  9,000  =  132,840  inch  Ibs.  Dividing  this  by  16,000  we  obtain 
8.3  for  section  modulus,  which  calls  for  2  —  [s  6"x8#,  but  to  obtain 
rigidity  2  —  [s  7"x9.75#  will  be  used.  The  stress  due  to  wind  load  is 
not  considered  as  it  does  not  exceed  50  per  cent  of  the  live-  and  dead-load 
stresses.  This  is  in  accordance  with  practice. 

For  the  maximum  end  shear  on  the  beam  we  have 

Dead  load  =  860    x  7.5  =  6^450  Ibs. 

Live    load=   62.5x7.5=    469  Ibs. 

6,9  19  Ibs. 

Beams  8.  For  the  load  on  each  of  these  beams  we  have  the  fol- 
lowing: 


DESIGN  OF  BUILDINGS 


63$ 


Concrete  slab  =  50x5  =  250  Ibs.  per  ft.  of  beam. 
Concrete  around  beam  =  120  Ibs.  per  ft.  of  beam. 
Beam  =  21  Ibs.  per  ft.  of  beam. 

Ceiling      =10x5  =50  Ibs.  per  ft.  of  beam. 

Live  load  =  25x5  =125  Ibs.  per  ft.  of  beam. 

Total  =  566  Ibs.  per  ft.  of  beam. 

Considering  the  beams  fixed  to  the  columns  by  brackets  we  have  TVX  566  X 
122x  12  =  81,504  inch  Ibs.  for  the  maximum  moment  at  the  ends  of  the 
beam  due  to  dead  and  live  load.  The  moment  due  to  wind  load  is  539  X 
72  =  38,808  inch  Ibs.,  which  is  less  than  50  per  cent  of  the  live-  and  dead- 
load  moment  and  hence  will  not  be  considered. 

Dividing  the  dead-  and  live-load  moment  by  16,000,  we  obtain 
81,504 -r  16,000  =  5.1  for  the  section  modulus,  which  calls  for  a  6"  x  12.25* 
I,  but  for  the  sake  of  rigidity  and  also  to  allow  for  rivet  holes  at  the 
brackets  we  will  use  1 — I  7"xl5#  for  each  beam. 

For  the  maximum  end  shear  on  the  beam  we  have 

Dead  load  =  441  x  7.5  =  3,307  Ibs. 

Live    load  =  125  x  7.5  =  _93^1bs. 

4,244  Ibs. 

Beams  4-  There  will  be  two  concentrations  on  each  of  these  beams, 
each  of  which  we  will  assume  is  equal  to  twice  the  maximum  end  shear  on 
beams  1.  The  concrete  around  the  beam  will  be  assumed  to  weigh  120 
Ibs.  per  ft.  of  beam  and  the  beam  itself  will  be  assumed  to  weigh  21  Ibs. 
The  loading  will  then  be  as  indicated  in  Fig.  447.  These  loads  include 
both  live  and  dead  load. 

Considering  the  beam  to  be  fixed  we  have  TV  X  141  x  122  X  12  =  20,304 
inch  Ibs.  for  the  maximum  moment  due  to  the  uniform  load.  A;  =  3. 5/12  = 
0.29  for  one  load  and  8.5/12  = 
0.71  for  the  other.  Then  substi- 
tuting in  (7)  of  Art.  69  we  obtain 
229,120  inch  Ibs.  for  the  maxi- 
mum moment  due  to  the  concen- 
trations. This  moment  occurs  at 
the  end  of  the  beam. 

Adding  together  the  two  mo- 
ments   we    obtain    20,304  +  229,-  Fig  447 
120  =  249,424    inch    Ibs.    for    the 

total  maximum  bending  moment  on  the  beam  due  to  dead  and  live  load. 
Dividing  this  by  16,000  Ibs.  we  obtain  15.5  for  the  section  modulus  which 
calls  for  a  9"x21#  I,  which  will  be  used  for  each  of  these  beams. 

For  the  maximum  end  shear  on  these  beams  we  have  7,724  + 141 X 
7.5  =  8,781  Ibs.     The  wind  stress  is  so  small  that  it  is  negligible. 
Beams  5.     The  dead  load  on  each  of  these  beams  is  as  follows : 

Fire  wall  (same  as  2)  =560  Ibs.  per  ft.  of  beam. 

Concrete  around  beam  =  150  Ibs.  per  ft.  of  beam. 

Beam  (2  channels)        =    30  Ibs.  per  ft.  of  beam. 

740  Ibs.  per  ft.  of  beam. 


\ 

s 

^ 

I 

N 

35' 

5' 

3.5' 

15' 

12' 

1.5' 

is' 

640  STRUCTURAL  ENGINEERING 

For  the  maximum  moment  on  the  beam  due  to  this  load,  considering 
the  beam  fixed,  we  have  TVx740xl22x  12  =  106,560  inch  Ibs. 

The  moment  due  to  the  two  concentrated  loads  will  be  one-half  of 
that  found  for  beams  4  or  229,120  H-  2  =  114,560  inch  Ibs.  Adding  the  two 
above  moments  together  we  have 

106,560  + 114,560  =  221,120  inch  Ibs. 

for  the  total  maximum  moment  on  the  beam.     Dividing  this  by  16,000  we 
obtain  13.8,  which  calls  for  2 — 8"x  11.25*  [s,  which  will  be  used. 

Beams  6  to  12.  For  the  sake  of  uniformity  beams  6  will  be  the  same 
size  as  beams  4,  although  the  moment  does  not  require  such  a  large  sec- 
tion. Likewise,  beams  7  will  be  the  same  size  as  beams  3 ;  beams  8,  9, 
and  10  will  be  the  same  size  as  beams  1 ;  and  beams  11  and  12  will  be  the 
same  size  as  beams  5. 

288.    Designing  of  the  Tenth-Story  Columns  — 
Load  on  Column  Cl 
Dead  load 

From  beams  2  6,450x2    =12,900  Ibs. 

From  beams  1  2,925x2    =    5,850  Ibs. 

From  beams  4  141  x  7£  =    1,058  Ibs. 

19^708"  Ibs. 

Live  load  =  25  x  15  x  7J  =   2,812  Ibs. 

Total  =  22,520  Ibs. 
Load  on  Column  C2 
Dead  load 

From  beams  2  =    6,450  Ibs. 

From  beams  1  =    2,925  Ibs. 

From  beams  5  =  740x7J  =   6,450  Ibs. 

15,825  Ibs. 

Live  load  =  25  x  7|  x  7J  =    1,406  Ibs. 

Total  =  17,231  Ibs. 

Load  on  Column  C3 
Dead  load 

From  beams  5  =  6,450x2  =12,900  Ibs. 

From  beams  1  =  2,925x2  =    5,850  Ibs. 

From  beams  3  =   3,307  Ibs. 

22,057  Ibs. 

Live  load  =  25  x  15  x  7J  =   2,812  Ibs. 

Total  =  24,869  Ibs. 

Load  on  Column  C4 
Dead  load 

From  beams  1  =  2,925x4  =11,700  Ibs. 

From  beams  4  =  1,058x2  =    2,116  Ibs. 

From  beams  3  =  3.307x2  =   6,614  Ibs. 

20,430  Ibs. 

Live  load  =  25  x  15  x  15  =   5,625  Ibs. 

Total  =  26,055  Ibs. 


DESIGN    OF  BUILDINGS  641 

Load  on  Column  C5 
Dead  load 

From  beams  1  =  5,850  Ibs. 

From  beams  3  =  3,307  Ibs. 

From  beams  7  =  2,200  Ibs. 

From  beams  8,  9  and  10  =  3,900  Ibs. 

From  beams  6  =   1,000  Ibs. 

16,257  Ibs. 

Live  load  =  25  X  12|  X  15  =  4,687  Ibs. 

Total  =  20,944  Ibs. 

Load  on  Columns  C6.  Total  =  about  f  of  load  on  column  C3  =  16,600 
Ibs. 

Load  on  Column  C7 
Dead  load 

From  beams  2  =   6,450  Ibs. 

From  beams  11  =  6,020  Ibs. 

From  beams  3  =  3,307  Ibs. 

From  beams  1  =  5,850  Ibs. 

From  beams  8  and  9  =  2,600  Ibs. 

24,227  Ibs. 

Live  load  =  4,200  Ibs. 

Total  =  28,427  Ibs. 

Load  on  Column  C7'.  The  load  on  this  column  is  about  equal  to  the  load 
on  column  C7  plus  10,000  Ibs.  for  elevator,  which  makes  a  total  load  of  about 
38,000  Ibs. 

Load  on  Column  C2'.  This  load  on  this  column  is  about  equal  to  the 
load  on  column  C2  plus  10,000  Ibs.  for  elevator,  which  makes  a  total  load  of 
about  27,000  Ibs. 

The  direct  wind  loads  on  these  columns  are  negligible,  the  maximum 
being  only  about  367  Ibs. 

The  maximum  shear  on  these  columns  due  to  wind  is  1,194  Ibs.  in  one 
direction  and  554  Ibs.  in  the  other  direction  (see  Figs.  443  and  445).  Then 
for  the  maximum  bending  moments  on  the  columns,  we  have  1,194X6X12 
=  85,968  inch  Ibs.  and  554X6X12  =  39,888  inch  Ibs.,  respectively. 

The  columns  should  be  designed  to  resist  this  bending  in  addition  to  the 
direct  load  on  them.     However,  the  unit  stress  for  the  combined  loading  can 
be  safely  increased  50  per  cent  over  that  obtained  by  the  formula,  16,000 
-707,/r. 

Let  us  assume  the  following  section: 

4-Ls  4"x3"xTV'  =   8.36°" 
1-web      12"  x'  =   3.75n" 


The  Max.  7  =  294.6    and  the  Min.  7  =  30.2 
The  Max.  r=      4.91  and  the  Min.  r=    1.58 

Columns   C4   in   the   center   of   the   building   will   be   subjected   to   the 


642  STRUCTURAL  ENGINEERING 

greatest  bending  in  the  two  directions.     For  the  fiber  stresses  we  have 

/=  39,888  x  4.1  +  30.2  =  5,415  Ibs. 
and 

f  =  85,968  x  6.2  +  294.6  =  1,809  Ibs. 

For  the  allowable  direct  stress,  we  have  16,000-  70  x  144-=-  1.58  = 
9,620  Ibs.  per  sq.  in. 

Dividing  the  area  of  cross-section  of  the  column  into  the  load  on  the 
column  we  have  26,055  -r  12.11  =  2,150  Ibs.  per  sq.  in.  for  the  direct  stress. 
Adding  this  to  the  maximum  wind  stress  we  have  5,415  +  2,150  =  7,565  Ibs. 
per  sq.  in.,  which  is  quite  low,  so  we  will  assume  a  lighter  section. 

Let  us  try 

4—  Ls   3i"x2J"xi"  =  5.76n" 
1—  web  "       12"x"  = 


8.76n" 

Max.  7  =  213.5,    Min.  7=15.9 
Max.  r=      4.94,  Min.  r=    1.35 

For  the  allowable  direct  stress  we  have  16,000-  70  x  144-=-  1.35  = 
8,530  Ibs.  per  sq.  in.  For  the  unit  stress  due  to  cross  bending  we  have 
39,888  x  3.62  -r  15.9  =  9,080  Ibs. 

For  the  stress  due  to  direct  load  we  have  26,055  -r  8.76  =  2,970  Ibs. 
per  sq.  in.,  which  is  low.  Adding  together  the  stresses  due  to  direct  load 
and  cross  bending  we  have  2,970  +  9,080  =  12,050  Ibs.  per  sq.  in.  The 
allowable  is  8,530x1.5  =  12,795  Ibs.  per  sq.  in.  for  combined  loading, 
which  is  very  near  the  actual  stress  and  hence  the  last  assumed  section 
can  be  used.  This  section  is  about  the  minimum  that  would  be  used,  so 
all  columns  in  the  tenth  story  will  have  this  section. 

289.  Designing  of  the  Tenth-Floor  Framing.  —  The  framing  in 
this  floor  will  be  as  indicated  in  Fig.  448. 

Beams  1.  For  the  dead  load  on  each  of  these  beams  we  have  the 
following  : 

4"  concrete  slab  =  50  x  5  =  250  Ibs.  per  ft.  of  beam. 
Plaster  =5x5=    25  Ibs.  per  ft.  of  beam. 

Concrete  around  beam     =    75  Ibs.  per  ft.  of  beam. 
Beam  =    18  Ibs.  per  ft.  of  beam. 

Total  =  368  Ibs.  per  ft.  of  beam. 

_  2 

For  the  maximum  moment  due  to  dead  load  we  have  ^  x  368  x  15  x!2  = 
124,200  inch_lbs.  For  the  maximum  moment  due  to  live  load  we  have 
-|  (50x5)  x!52x  12  =  84,375  inch  Ibs.  Then  for  the  total  maximum  mo- 
ment we  have  124,200+84,375  =  208,575  inch  Ibs.  Dividing  this  by  16,000 
we  obtain  13.04,  which  calls  for  an  8"  x  18#  I,  which  will  be  used. 
For  maximum  end  shear  on  the  beam  we  have 

Dead  load  =  368  x  7^  =  2,760  Ibs. 

Live    load  =  250  x  7£  =  1,875  Ibs. 

Total  =  4,635  Ibs. 

Beams  2.     The  dead  load  on  each  of  these  beams  consists  oi  a  story 


DESIGN  OF  BUILDINGS 


643 


of  12-in.  curtain  wall,  a  strip  of  concrete  floor  2.5  ft.  wide,  and  the  weight 
of  the  beam  and  concrete  encasing  it. 
For  the  weight  of  the  wall  we  have 

8"  hollow  tile=   40  x  0.66  =  26.4  Ibs.  per  sq.  ft.  of  wall. 

4"  brick  face    =  140  x  0.33  =  46.2  Ibs.  per  sq.  ft.  of  wall. 

Total  =  72.6  Ibs.  per  sq.  ft.  of  wall. 

The  curtain  wall  in  each  bay  will  have  two  openings  for  windows, 
as  shown  at  (a),  Fig.  449.     The  load  on  the  beam  from  this  wall  will  be 


60-0' 


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Floor  Framing. 

Fig.  448 


as  shown  at  (6),  but,  as  the  load  near  the  ends  of  the  beam  is  supported 
directly  upon  the  brackets,  the  load  for  computing  the  stress  on  the  beam 
can  be  taken  as  indicated  at  (c)  . 


644 


STRUCTUKAL  ENGINEERING 


Then  for  the  maximum  moment  on  the  beam  due  to  the  curtain  wall, 
we  have 

A+  182  x  II2  x  12  =  22,022  inch  Ibs. 


-x  2,504x11  x  12  =  41,318  inch  Ibs. 
Total  =63,340  inch  Ibs. 


2' 


2' 


(a) 


Tig.  449 

For  the  remainder  of  the  dead  load  we  have 

Concrete    slab  =  50  x  2.5  =  125  Ibs.  per  ft.  of  beam. 

Beam   (2—  [s)  =    30  Ibs.  per  ft.  of  beam. 

Concrete  around  beam  =150  Ibs.  per  ft.  of  beam. 

Total  ="305  Ibs.  per  ft.  of  beam. 


DESIGN  OF  BUILDINGS  645 

For  the  moment  on  the  beam  due  to  this  load  we  have 

~  x  305  x  IT*  x  12  =  36,905  inch  Ibs. 
12 

For  the  live  load  on  the  beam  we  have  50  x  2.5  — 125  Ibs.  per  ft.  of 
beam.  Then  for  the  live-load  moment  on  the  beam  we  have 

-i  x  125  x  IT'  x  12  =  15,125  inch  Ibs. 
12 

Adding  together  all  of  the  above  moments  we  have 

63,340  +  36,905  +  15,125  =  115,370  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  beam  due  to  dead  and  live 
load. 

The  maximum  shear  on  the  beam  due  to  wind  load  is  1,162 -r  2  =  576 
Ibs.  (see  Pig.  440).  Then  for  the  moment  on  the  beam  due  to  wind  load 
we  have  576x66  =  38,016  inch  Ibs.,  which  is  less  than  50  per  cent  of  the 
live-  and  dead-load  moment  and  hence  it  will  be  ignored.  Dividing  the 
dead-  and  live-load  moment  by  16,000  we  have  115,370 -r  16,000  =  7.2  for 
the  section  modulus,  which  calls  for  2 — [s  6"  x  8*,  but  for  the  sake  of 
rigidity  we  will  use  2— [s  7"x9.75#. 

,,  fDead  load  =  6,504  Ibs. 

Maximum  end  shear ^  T.         -,     j        ^orv  ,, 

l^Live    load  =    937  Ibs. 

7,441  Ibs. 

Beams  3.  For  the  dead  load  on  each  of  these  beams  we  have  the 
following : 

Partition  =  40  x  0.33  x  12  =  160  Ibs.  per  ft.  of  beam. 
Concrete  slab  =  50x5  =  250  Ibs.  per  ft.  of  beam. 
Beam  =  21  Ibs.  per  ft.  of  beam. 

Concrete  around  beam  =  150  Ibs.  per  ft.  of  beam. 
Plaster  =  5x5  =  25  Ibs.  per  ft.  of  beam. 

636  Ibs.  per  ft.  of  beam. 

Then  for  the  maximum  moment  on  the  beam  due  to  dead  load,  we 
have  TV  x  636  xll2x  12  =  76,956  inch  Ibs.  The  live  load  is  50x5  =  250 
Ibs.  per  ft.  of  beam.  Then  for  the  moment  on  the  beam  due  to  live  load, 
we  have  ^  x  250  xll*  x  12  =  30,250  inch  Ibs.  Adding  together  the  dead- 
and  live-load  moments  we  obtain  107,206  inch  Ibs.  for  the  total  maximum 
moment  due  to  dead  and  live  load.  The  maximum  shear  on  these  beams 
due  to  wind  load,  as  given  in  Fig.  440,  is  1,152  Ibs.  So,  for  the  maximum 
bending  moment  on  the  beams  due  to  wind,  we  have  1,152x66  =  76,032 
inch  Ibs.  Now  subtracting  from  this  50  per  cent  of  the  live-  and  dead- 
load  moment  we  obtain  76,032  -  53,603  =  22,429  inch  Ibs.,  which  must  be 
added  to  the  maximum  dead-  and  live-load  moment.  So  we  have 

107,205  +  22,429  =  129,635  inch  Ibs. 
bending  moment  which  the  beam  must  be  designed  to  resist.     Dividing 


646 


STRUCTUKAL  ENGINEERING 


this  by  16,000  we  obtain  8.1  for  the  section  modulus  which  calls  for  a 
7"xl5#  I,  but  as  beams  1  are  8-in.  beams  we  will  make  these  the  same 
size;  that  is,  for  each  we  will  use  1 — I  8"  x  18*.  This  excess  size  also 
provides  for  rivet  holes  cut  out  at  the  bracket. 


Maximum  end  shear 


\ 

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2' 

II  ' 

z' 

15' 

Dead  load  =  636  x  7.5  =  4,770  Ibs. 
Live    load  =1,875  Ibs. 

6,645  Ibs. 

Beams  4-  The  maximum  dead  load  consists  of  a  4-in.  partition, 
two  concentrations  from  beams  1,  the  weight  of  the  beam,  and  the  concrete 
encasing  it.  For  the  uniform  dead  load  on  the  beam  we  have  the 
following: 

Partition   (same  as  3)  =  160  Ibs.  per  ft.  of  beam. 

Beam  =    25  Ibs.  per  ft.  of  beam. 

Concrete  around  beam  =  150  Ibs.  per  ft.  of  beam. 

Total  =  335  Ibs.  per  ft.  of  beam. 

For  each  concentration  due  to  dead  load  we  have  2,760x2  =  5,520 

Ibs.  For  each  live-load  con- 
centration we  have  1,875  x  2  = 
3,750  Ibs.  Then  for  the  to- 
tal of  each  concentration  we 
have  5,520  +  3,750  =  9,270  Ibs. 
The  loading  will  then  be  as 
indicated  in  Fig.  450.  For 
the  maximum  moment  (con- 
sidering the  beam  fixed)  due 
Fig.  450  to  the  uniform  load  we  have 

—  x  335  x  IT2  x  12  =  40,535  inch  Ibs. 

For  the  maximum  moment  due  to  the  concentrated  loads  we  have 

[9,270 x  11  (2  x  O272  -  (m*  - 0.27)  +  9,270 x  11  (2 xOT?  - (XTS* 

-0.73)]  12  =  240,800  inch  Ibs. 

Now  adding  together  the  above  moments  we  obtain 
40,535  +  240,800  =  281,335  inch  Ibs. 

for  the  total  maximum  dead-  and  live-load  moment.  The  maximum  shear 
on  these  beams  due  to  wind  load,  as  seen  from  Fig.  445,  is  731  Ibs.  So, 
for  the  maximum  moment  on  the  beams  due  to  wind  load  we  have  731  x 
66  =  48,246  inch  Ibs.,  which  is  less  than  50  per  cent  of  the  combined  dead- 
and  live-load  moment  and  hence  will  be  ignored. 

Then  for  the  required  section  modulus  we  have  281,335  +  16,000  = 
17.57,  which  calls  for  1— I  9"  x  21  #,  which  will  be  used. 

For  the  maximum  end  shear  on  the  beam  we  have 

Dead  load  =    8,032  Ibs. 

Live    load=   3,750  Ibs. 

Total  =11,782  Ibs. 


DESIGN  OF  BUILDINGS  647 

Beams  5.  The  dead  load  supported  by  each  of  these  beams  consists 
of  a  story  of  12-in.  curtain  wall,  two  concentrations  from  beams  1,  and  a 
uniform  load  which  consists  of  the  weight  of  the  beam  and  the  concrete 
encasing  it. 

The  bending  moment  due  to  the  curtain  wall  =  63,340  inch  Ibs.,  the 
same  as  for  beams  2. 

The  bending  moment  due  to  the  concentrations  =  120,400  inch  Ibs., 
one-half  of  that  for  beams  4. 

For  the  uniform  dead  load  we  have  the  following: 

Beam   (2 — [s)  =    25  Ibs.  per  ft.  of  beam. 

Concrete  around  beam  =  150  Ibs.  per  ft.  of  beam. 

Total  =  175  Ibs.  per  ft.  of  beam. 

For  the  maximum  bending  moment  due  to  this  load  we  have 

~  x  175  x  IT*  x  12  =  21,175  inch  Ibs. 
±2 

Adding  together  all  of  the  above  moments  we  obtain 

63,340  + 120,400  +  21,175  =  204,915  inch  Ibs. 

for  the  total  maximum  moment  due  to  dead  and  live  load.  Dividing  this 
by  16,000  we  obtain  12.8  for  the  section  modulus  which  calls  for  2 — [s 
8"x  11.25*,  which  will  be  used. 

The  wind  stress  is  ignored  as  it  is  less  than  50  per  cent  of  the  live- 
and  dead-load  stresses  combined. 

For  the  maximum  end  shear  on  the  beam  we  have  the  following: 

Dead  load  =    8,290  Ibs. 

Live    load  =    1,875  Ibs. 

10,165  Ibs. 

Beams  5' .  The  loading  on  these  beams  is  exactly  the  same  as  on 
beams  5,  except  for  the  wall,  which  is  solid,  no  openings  for  windows  at 
all.  This  wall,  as  determined  for  beams  2,  weighs  72.6  Ibs.  per  sq.  ft. 
Then,  ignoring  arch  action,  we  have  72.6x12  =  871  Ibs.,  per  ft.  of  beam. 
For  the  moment  due  to  this  load  we  have 

^r  x  871  x  II2  x  12  =  105,391  inch  Ibs. 
12 

Now,  adding  this  to  the  other  moments  found  for  beams  5,  we  obtain 
105,391  +  120,400  +  21,175  =  246,966  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  beam.  Dividing  this  by 
16,000  we  obtain  15.4  for  the  section  modulus,  which  calls  for  2 — [s  8"  x 
11.25* — the  same  as  used  for  beams  5. 

For  the  maximum  end  shear  we  have  the  following: 

Dead  load  =  10,604  Ibs. 

Live    load  =  J^^  Ibs. 

12,479  Ibs. 


648  STEUCTURAL  ENGINEERING 

Beams  6,  7,  8,  and  9.  For  the  sake  of  uniformity  beams  6  will  be 
the  same  size  as  beams  3,  although  the  moment  does  not  require  such  a 
heavy  section.  Likewise,  beams  7,  8,  and  9  will  be  the  same  size  as 
beams  1. 

Beams  10.     For  the  dead  load  on  these  beams  we  have  the  following: 

Partition  =  160  Ibs.  per  ft.  of  beam. 

Plaster  =    25  Ibs.  per  ft.  of  beam. 

Concrete  slab  =  250  Ibs.  per  ft.  of  beam. 

Beam  =    25  Ibs.  per  ft.  of  beam. 

Concrete  around  beam-    75  Ibs.  per  ft.  of  beam. 
535  Ibs.  per  ft.  of  beam. 

Then  for  the  dead-load  moment  we  have 

i  x  535  x  IS*  x  12  =  180,562  inch  Ibs. 
8 

The  maximum  moment  due  to  live  load  is  84,375  inch  Ibs.  the  same  as 
for  beams  1.  Adding,  we  have 

180,562  +  84,375  =  264,937  inch  Ibs. 

for  the  total  maximum  bending  moment  on  the  beam.     Dividing  this  by 
16,000  we  obtain  16.56  for  the  required  section  modulus,  which  calls  for 
1 — I  8"x23#.     Metal  would  be  saved  by  using  the  9"x21#  I  but  it  is 
desirable  to  have  beams  10  of  the  same  depth  as  beams  1. 
Maximum  end  shear  is  as  follows: 

Dead  load  =  3,960  Ibs. 

Live    load  =  1,875  Ibs. 

5,835  Ibs. 

Beams  1'.  The  loading  on  these  beams  is  exactly  the  same  as  that 
on  beams  1,  except  a  partition  is  supported  at  mid-span.  For  this  con- 
centration we  have  160x5  =  800  Ibs.  This  produces  a  moment  of  400  x 
7.5  x  12  =  36,000  inch  Ibs.  Adding  this  to  the  maximum  moment  on  beams 
1  we  have 

208,575  +  36,000  =  244,575  inch  Ibs. 

for  the  total  maximum  moment  on  beams  V '.  Dividing  this  by  16,000  we 
obtain  15.3  for  the  required  section  modulus,  which  calls  for  an  8"  x  20.5* 
I,  which  will  be  used. 

Beams  3'  and  4' '•  Beams  3'  support  less  load  from  partitions  than 
beams  3,  but  the  difference  is  not  sufficient  to  change  the  section,  so  beams 
3'  will  be  the  same  size  as  beams  3.  Beams  4'  support  no  partitions  and 
consequently  could  be  lighter  than  beams  4,  but  for  the  sake  of  uniformity 
they  will  be  the  same  size  as  beams  4. 

The  floor  framing  for  the  other  floors  down  to  the  first  floor  would 
be  the  same  as  the  tenth  floor  if  it  were  not  for  the  wind  stresses.  How- 
ever, the  work  of  designing  the  framing  in  these  floors  is  exactly  the  same 
as  shown  above  for  the  tenth  floor,  and  hence  the  method  of  procedure  is 
fully  shown  above. 

As  an  example,  let  us  consider  beams  4  in  the  fourth  floor.     The 


DESIGN  OF  BUILDINGS  649 

sum  of  the  dead-  and  live-load  moments  is  231,335  inch  Ibs.,  as  found 
above.  The  maximum  shear  on  the  beams  due  to  wind,  as  given  in  Fig. 
445,  is  4,023  Ibs.  Then  for  the  maximum  moment  due  to  wind  load  we 
have 

4,023x5.5x12  =  265,518  inch  Ibs. 

Subtracting  50  per  cent  of  the  dead-  and  live-load  moment  from  this,  we 
have  265,518-140,667  =  124,851  inch  Ibs.  So  we  have 

281,335  +  124,851  =  406,186  inch  Ibs. 

for  the  moment  which  the  beam  must  be  designed  to  resist.  Dividing  this 
by  16,000  we  obtain  25.4  for  the  section  modulus,  which  calls  for  a  10"  x 
30#  I,  which  will  be  used. 

The  floor  framing  on  the  first  floor  is  designed  to  support  a  live  load 
of  100  Ibs.  per  sq.  ft.  of  floor.  The  work  of  designing  this  framing  is 
the  same  as  for  the  other  floors. 

290.  Designing  of  the  Ninth-Story  Columns.— As  an  example, 
let  us  consider  column  C4.  (See  Fig.  448.) 

For  the  maximum  direct  load  on  this  column  we  have  the  following: 

From  roof  (constant)  =26,055  Ibs. 

From  beams  3  =  4,770x2          =    9,540  Ibs. 

From  beams  4  =  8,032x2  =16,064  Ibs.  |*36,854  Ibs. 

Live  load          =  50  x  15  x  15     =  11,250  Ibs.  J 

Weight  of  column  and  concrete 

encasing  it  =    2,400  Ibs. 

Total  =  65,309  Ibs. 

The  maximum  direct  wind  load  equals  514  Ibs.  (negligible)  (see  Fig. 
443).  The  maximum  shears  on  the  column  equal  2,626  Ibs.  and  1,218 
Ibs.  (see  Figs.  443  and  445).  Then  for  the  maximum  moments  on  the 
column  at  the  brackets  we  have  2,626  x  (6-2)  12  =  126,048  inch  Ibs.,  and 
1,218  x  48  =  58,464  inch  Ibs. 

Let  us  assume  the  following  section: 

4—Ls  4x3xT5e  =   8.36°" 

1— web  12  xA  =    3.75n" 

Total  =  12.11n" 

Max.  7  =  294.6     and  Min.  7  =  30.2 
Max.  r=      4.91  and  Min.  r-    1.58 

For  the  allowable  unit  stress  for  dead  and  live  loads  we  have  16,000  - 
70  x  144  -r  1.58  =  9,620  Ibs.  Dividing  this  into  the  load  we  have  65,309  + 
9,620  =  6.79n//  for  the  required  area  of  cross-section.  For  the  actual 
direct  stress  we  have  65,309  +  12.11  =  5,390  Ibs.  per  sq.  in. 

For  the  maximum  unit  stresses  due  to  cross  bending  (from  the  wind 
load)  we  have 

/  =  126,048  x  6  +  294.6  =  2,568  Ibs. 

and 

/'  =  58,464  x  4.15  -r  30.2  =  8,038  Ibs. 


650  STRUCTURAL  ENGINEERING 

Adding  the  maximum  of  these  to  the  direct  stress  we  obtain 
8,038  +  5,390  =  13,428  Ibs. 

for  the  maximum  combined  unit  stress  on  the  column.  For  the  allowable 
unit  stress  for  the  combined  dead,  live,  and  wind  loads  we  have  9,620  x 
1.5  =  14,430  Ibs.  From  this  it  is  seen  that  the  above  assumed  section  is 
about  the  correct  size  and  hence  could  be  used. 

The  other  columns  in  the  ninth  story  are  designed  in  the  same 
manner  as  shown  for  C4.  Unless  the  loads  are  quite  different  the  columns 
throughout  the  story  would  be  of  the  same  section — using  the  maximum 
section  required. 

After  the  columns  in  the  tenth  and  ninth  stories  are  designed,  as  ex- 
plained above,  the  columns  in  the  eighth  can  be  designed  in  the  same 
manner,  and  then  the  columns  in  the  seventh  story ;  and  so  on  down  to  the 
basement  columns,  using  in  each  case  the  loads  found  at  the  stories  above. 
As  an  example,  let  us  consider  column  C4  in  the  first  story. 

For  the  direct  load  on  the  column  (see  loads  on  C4  given  above),  we 
have  the  following: 

From  the  roof  =    26,055  Ibs. 

From  10th,  9th  ...  2nd  floors  =  3 6,854  x  9  =331,686  Ibs. 

Weight  of  column  above  and  concrete  encasing  it  =    37,800  Ibs. 

395,541  Ibs. 

For  the  direct  wind  load  on  the  column  we  will  take  15,006  Ibs.  (see 
Fig.  443). 

For  the  maximum  moments  due  to  wind  loads  we  have 

14,442  x  (9  -  3)  12  =  1,039,824  inch  Ibs. 
and 

6,700x72-482,400  inch  Ibs. 

Let  us  assume  the  section  shown  in  Fig.  451  which 
is  composed  of  the  following  section: 

2— [s  15"x40#  =  23.52n" 

2— [s  12"x25#  =  14.70n" 
Total  =  38.22°" 

Max.  I  =  1,389,       Min.'  7=711 
Fi&.  451  Max.  r-         6.02,  Min.  r=      4.3 

For  the  allowable  unit  stress  for  live  and  dead  load  we  have 
16,000-70x216-^-4.3  =  12,470  Ibs. 

Dividing  this  into  the  sum  of  the  dead  and  live-load  stresses  we  obtain 
395,541 -r  12,470  =  31.7n//  for  the  required  section  for  dead  and  live  load 
alone. 

For  the  total  direct  stress — dead,  live,  and  wind  loads  combined — we 
have 

395,541  +  15,006  =  410,547  Ibs. 


652  STRUCTURAL    ENGINEERING 

Dividing  this  by  the  area  of  the  cross-section  we  obtain  410,547  -5-  38.22  = 
10,740  Ibs.  for  the  actual-  direct  stress"  on  the  column  due  to  dead,  live,  and 
wind  loads  combined.  For  the  unit  stresses  due  to  cross  bending  we  have 

/•=  1,039,824X9.4-:-  1,389  =  7,035  Ibs. 
and 

/'  =  482,400X7.5-J-71  1  =  5,090  Ibs. 

Then  for  the  maximum  unit  stress  we  have 

10,740+7,035  =  17,775  Ibs. 

For  the  allowable  stress  we  have  12,470X1.5  =  18,705  Ibs.  for  combined 
dead,  live,  and  wind  loads.  From  this  it  is  seen  that  the  assumed  section  is 
about  correct  and  hence  could  be  used. 

The  other  columns  throughout  the  building  are  designed  in  the  same 
manner. 

291.  Calculation  of  Details.—  A  fair  idea  of  the  details  of  the 
building  can  be  obtained  from  Fig.  452.  As  is  seen,  knee  braces  are  used 
throughout  instead  of  the  ordinary  brackets.  The  connections  of  the 
beams  to  the  columns  in  all  cases  are  sufficient  to  develop  the  bending 
moment. 

As  an  example,  let  us  consider  the  details  connecting  beam  3  to  column 
4,  as  shown  at  (a),  Fig.  452.  The  moment  at  0,  as  given  in  Art.  289,  is  129,635 
inch  Ibs.  and  the  end  shear  is  6,645  Ibs.,  which  we  will  assume  applied  at  0. 
Then,  for  the  moment  about  the  center  of  the  column  (see  Art.  66)  ,  we  have 
129,635+6,645X18  =  249,245  inch  Ibs.  Then  dividing  this  by  the  arm 
b  (15")  we  obtain  249,245  -r-  15  =  16,616  Ibs.  for  the  stress  in  the  knee  brace, 
provided  there  were  but  one  brace,  but,  as  there  are  two  holding  the  beam, 
stress  in  each  knee  brace  will  be  16,616  -r-  2  =  8,308  Ibs.  As  is  seen,  the  stress 
in  the  knee  brace  causes  shear  and  tension  on  the  rivets  connecting  it  to  the 
beam  and  column.  So,  as  a  compromise,  we  will  compute  the  rivets  to 
take  the  total  stress  in  each  knee  brace.  Then  we  have  8,308  -i-  4,420 
=  1.88,  say  2—  -f"  field  rivets  required  to  connect  each  knee  brace  to 
the  beam;  also  to  the  column.  As  is  seen,  4  are  used  in  order  to  obtain  good 
connections.  We  will  next  determine  the  number  of  rivets  required  to 
connect  the  end  of  the  beam  to  the  column.  Taking  moments  about  0  we 
have 


129,635 

from  which  we  obtain 

R=  7,200  Ibs. 

for  the  force  exerted  along  the  column  by  the  beam.  So  we  have  required 
7,200  H-  4,420  =  1.6,  say  2-  f"  field  rivets.  As  is  seen,  4  are  used.  All  other 
connections  of  the  beams  to  the  columns  throughout  the  building  can  be  cal- 
culated in  the  same  manner. 

The  other  details  throughout  the  building  are  readily  calculated,  provided 
the  work  previously  given  in  this  book  is  understood. 


For  one  re//  on/y. 


s  Fo//owed  '  by4OOO  Ibs.per  ft.. 
CLASS  -£_-4O 


TABLE  A. 


/a*. 


ff- 


/03 


JS- 


232 


%- 


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7734  6793  5857  4999 


65_ 


68X5888  50 


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22/6/5/6 


208 


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59/9  5O484242 


35/4 


2851 


2421 


78. 


4JI8  339C 


2/55 


78 


V936  8336  6836  5436 


320  840  460 


8728 


7936.  65/8 


28582247 


/233 


559  884 


7668 


692856034398  3268 


2248  17O2 


'22 


8/8 


480 


300 


282 


529 


6708 


6018  4798  3678  2658 


/738 


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503 


230 


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30O 


547  859 


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5648 


5208  4O8B  3068  2/48 


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345 


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3834  4562  5366  &3 


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Areas  and  Weights  of  Bars  and  P/afes 

A  =  area  of  cross  -sect/on  in  sq.  inches 
T  Qul6-  /      W=  weight  in  pounds  per  /inea/ff.  at  490  pounds  per  cuff: 

WWh 

/n 
Mes 

Thickness 

Extreme 

Length 
/n 
Feef 

• 

/  " 

2 

£" 
/6 

i" 

%' 

r 

/f 

^ 

w 

A 

W 

A 

W 

A 

W 

A 

W 

A 

W 

/ 

O.2& 

O.S5 

O.3/3 

W6 

0.375 

/.28 

0.438 

/49 

V5W 

170 

C.563 

/.9? 

70 

1 

0.375 

/.28 

0469 

/.£9 

0563 

/92 

0.656 

Z?3 

O.750 

2.55 

0.844 

2#7 

it 

2 

O.438 

/.49 

054'/ 

/.86 

0656 

223 

O766 

260 

3875 

298 

0.934 

335 

if 

z 

0.500 

/.70 

0.625 

2/2 

0750 

255 

0875 

293 

/.OO 

£40 

//3 

383 

/t 

% 

0563 

/.9/ 

0703 

2.39 

0.&44 

237 

0.934 

3.35 

/./3 

3S3 

/27 

4.30 

it 

?4 

0625 

2./2 

O.7SI 

Z65 

0.933 

J./9 

/.09 

372 

/.25 

4.25 

/4/ 

47<$ 

///, 

3 

0.750 

Z.55 

0.938 

3./9 

//3 

333 

/.3I 

4.46 

/.50 

J/O 

/.69 

574 

n 

3 

O.87S 

2.98 

/.09 

372 

/3/ 

447 

/•53 

5.?0 

/.75 

5.95 

/.97 

670 

n 

4 

/OO 

3.40 

/?5 

42S 

/.50 

S/0 

/75 

5.9S 

ZOO 

6.80 

?.?5 

765 

n 

5 

/25 

4.25 

/.56 

S.3/ 

f.M 

6.30 

2/9 

744 

250 

8.5O 

ZS/ 

9.57 

it 

6 

/50 

S./O 

/.8ff 

6.38 

Z.Z5 

7.65" 

263 

8.93 

300 

/0.20 

338 

//4£ 

If 

7 

/75 

£95 

2/9 

7.44 

263 

8.93 

306 

/O4/ 

3.50 

//.9O 

3.94 

/3.39 

ti 

8 

2OO 

6£O 

25O 

3.5-0 

3.OO 

/020 

350 

//.90 

4OO 

/3£0 

4.50 

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n 

9 

225 

7.6S 

?.0/ 

9.56 

338 

i/48 

394 

/3.40 

450 

/5.3O 

S06 

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n 

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250 

8.50 

3./J 

/a  62 

375 

/?7£ 

4.38 

/4S8 

£00 

/7OO 

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n 

// 

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9.14 

344 

//.6£ 

4/3 

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4.8/ 

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/87O 

6./9 

2/02 

ii> 

£ 

300 

/0.20 

3.75 

/2.7S 

4.50 

/5.30 

5.25 

/7.05 

600 

ZO40 

6.75 

2295 

M 

/s 

325 

//.06 

4.06 

/3.8/ 

438 

/6.58 

£69 

/934 

650 

22./0 

7.3/ 

24J6 

n 

/4 

3.50 

//.90 

4.38 

/4S3 

J.Z5 

/7<36 

6./3 

20.82 

700 

2380 

7.88 

26.78 

H 

/5 

37S 

f?75 

4.69 

/S.94 

563 

&/<4 

6.56 

22.32 

7.50 

25.50 

844 

2870 

it 

/6 

4.00 

/3.60 

S.oo 

/7.0O 

6.00 

20.40 

700 

23.80 

£.00 

2770 

9.00 

30.60 

IV 

/7 

425 

/444 

£3/ 

/#06 

6.38 

2/6S 

744 

25.28 

850 

28.89 

9.56 

3252 

II' 

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450 

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5.63 

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6.7S 

2296 

7S8 

2679 

900 

3O.60 

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34.44 

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6.25 

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750 

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2975 

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3400 

//.25 

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6.68 

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9.63, 

3272 

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ii 

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20.40 

7.50 

25.52 

9.00 

3O.60 

/0.50 

3572 

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40.80 

/3.50 

45.92 

n 

26 

6.50 

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S./3 

27.62 

9.75 

33/6 

//.3£ 

3868 

/300 

4420 

M.63 

4973 

n 

28 

7OO 

Z3JO 

6.75 

2976 

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35.72 

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4/65 

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47.60 

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5356 

// 

30 

750 

25.50 

9.36 

3/.68 

//.2£~ 

38.78 

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4464 

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5/.00 

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n 

32 

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27.20 

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34.00 

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47.60 

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54.40 

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28.85 

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36./2 

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43.36 

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£7.7<3 

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// 

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900 

30.59 

//.2£ 

38.24 

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45.92 

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2025 

6868 

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32.32 

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Areas  and  Weights  of  Bors  and  Plates 

A  =  area  of  cross-  sect/on  in  sq.  inches 
IQ  D  I6~S     W*  weight  in  pounds  per  //nee/  ft.  at  4  9O  pounds  per  cu.  ft 

Width 
in 
Inches 

Thickness 

Exirfme 

Length 
in 
reef. 

f" 

2L" 
/6 

i" 

J3_" 
16 

7  ff 

J5_" 
16 

A 

W 

A 

W 

A 

W 

A 

W 

A 

W 

A 

W 

/ 

0.625 

2.13 

O.G88 

2.34 

0.75 

2.55 

O.QI3 

2.76 

0.88 

2.98 

0.94 

3./9 

7O 

2 

0.93$ 

3.19 

1.021 

3.51 

/J25 

3.83 

f.2/9 

4./4 

/.3/ 

4-.4t> 

/.<?-/ 

4.7ft 

// 

2S 

1.094 

3.72 

/.203 

4.09 

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4-.46 

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4.83 

/.53 

S.2/ 

/.64 

5.58 

// 

2 

/.Z5 

4,25 

1.375 

4.68 

£50 

5./0 

/.625 

5.53 

1.75 

£-.35 

/.88 

6-38 

// 

H 

J.40& 

4t78 

1.541 

5".2£ 

/.68ft 

5.74 

/.828 

6.72. 

/.97 

6.69 

2.// 

7-/7 

f/ 

z-k 

1.563 

5.31 

1.7/9 

5.84 

/.87S 

6,38 

2.03' 

6.5/ 

2./9 

7.4-4 

2.34 

7.97 

f/ 

3 

1.875 

6.38 

2.063 

7.O/ 

2.25 

7.65 

2.4-3 

829 

2.63 

ft.93 

2.81 

9.56 

// 

3^ 

2.188 

7.44 

2-4U 

8>J8 

2.615 

833 

2-B44 

9.67 

3.O6 

fO.4t 

3.28 

S/./6 

// 

4 

2.5 

$.£0 

2.75 

9.35 

3.00 

/a  20 

3.2.5 

I/.O5 

3.5-0 

//.96 

3.YS 

/2.7S 

n 

£ 

3.125 

10.63 

3438 

I/.69 

3.7S 

12.75 

4.063 

13.81 

4r38 

/4.88 

^.69 

/S.94 

if 

6 

3.75 

12-75 

4.125 

/4.03 

4-.5O 

15.30 

4.875 

16.58 

S.2S 

/7.8S 

S.63 

/9</3 

// 

7 

4.375 

14.88 

4.8  1  3 

16-36 

5.250 

17.85 

5.688 

1934 

6./3 

20.83 

6.^6 

22.31 

// 

g 

5.00 

I7.O 

5.50 

18.70 

6.00 

20.40 

6.50 

22.10 

7.00 

2330 

7.W 

25.  5 

n 

3 

5.62S 

19.13 

6.188 

21.04 

6.  75 

2235 

7.3/3 

2436 

7.88 

26.78 

8.4+ 

28.69 

n 

/O 

6.25 

21.25 

6S75 

23.38 

7.50 

25.50 

S-/25 

27.G3 

8.7S 

29.75 

9.38 

3/.S8 

// 

// 

6JB7S 

23.28 

7563 

25.71 

8.25 

28.05 

8338 

30.39 

9.63 

32.73 

/0.3/ 

35.06 

tt 

/'Z 

7.50 

25.50 

8.25 

28.05 

9.00 

30.60 

9.7S 

33.15 

'O.50 

35.70 

//25 

38.25 

fi 

/3 

8J3 

27.63 

8.94 

30.4 

9.75 

33.2 

10.56 

35.3 

/I38 

38.7 

/2./9 

41.4 

n 

t± 

8-75 

29.75 

9.63 

32.7 

IO.SO 

3^.7 

11.38 

38.7 

12.25 

41.7 

/3J3 

44.6 

// 

IS 

9.3$ 

31.88 

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INDEX 

PAGES 

Applied   forces,    definition   of 17 

Beams,  theoretical   treatment  of 51  to  105 

Beams,  shearing  stresses  on 51  to     53 

Beams,  bending  stresses   on 53  to     56 

Beams,  reaction   on 56 

Beams,  deflection    of 72  to  106 

Beams,  cantilever    75  to     80 

Beams,  simple    80  to     84 

Beams,  fixed  at  one  end  and  supported  at  other 86  to     90 

Beams,  fixed    90  to     94 

Beams,  overhanging    84  to     86 

Beams,  continuous    94  to  105 

Beam  bridges,  railroad 198  to  214 

Beam  bridges,  highways    540  to  546 

Bending  stress,  increment  of   64  to     65 

Bending  moment,  maximum  on  simple  beams     128  to  132 

Bending  moment,  maximum  on  simple  trusses     138  to  139 

Bridges  on  curves   571  to  575 

Buildings,  mill     583  to  624 

Buildings,  office     624  to  652 

Camber  of   trusses    518  to  532 

Cantilever   beams    75  to     80 

Castings    2 

Centrifugal  force,  stresses  due  to 571  to  575 

Center  of  gravity,  determination  of 44  to     47 

Columns,  theoretical  treatment  of 106  to  114 

Column,   formulas    106  to  109 

Continuous  beams,  theoretical  treatment   94  to  105 

Conventional  signs    11 

Cover  plates,  theoretical  length   181  to  183 

Dead  load,  railroad  bridges    197  to  198 

Dead  load,  highway  bridges    534 

Deflection  of  beams    72  to  106 

Deflection  of  trusses     518  to  532 

Design  of  I-beams  and  plate  girders,  theoretical    172  to  194 

Design  of  railroad  bridges  : 

Beam  bridges   198  to  214 

Deck  plate  girders    215  to  259 

Through  plate  girders    259  to  295 

Viaducts    295  to  324 

Truss  bridges   324  to  532 

Design  of  highway  bridges  : 

Beam  bridges    540  to  546 

Pony  trusses   546  to  552 

Through  Pratt  trusses   554  to  564 

Curved  chord  through  trusses   564  to  589 

Design  of  a  10  ft.  beam  bridge,  railroad    199  to  201 

Design  of  a  15  ft.  beam  bridge,  railroad    201  to  214 

Design  of  a  50  ft.  deck  plate  girder,  railroad   216  to  244 

Design  of  a  75  ft.  deck  plate  girder,  railroad   244  to  248 

Design  of  a  90  ft.  deck  plate  girder,  railroad 249  to  253 

Design  of  a  60  ft.  through  plate  girder,  railroad   260  to  286 

Design  of  a  150  ft.  Pratt  truss  bridge,   railroad    326  to  413 

667 


668  INDEX 

PAGES 

Design  of  a  225  ft.  curved  chord  truss  bridge,  railroad   426  to  473 

Design  of  buildings     583  to  652 

Distortion,  definition    of    19 

Distortion,  determination  of 20  to     22 

Drawing  instruments    9 

Drawings,  size  of    11 

Drawings,  shop    11 

Drawing  Room  Exercise  No.  1    13 

Drawing  Room  Exercise  No.  2   214 

Drawing  Room  Exercise  No.  3   259 

Drawing  Room  Exercise  No.  4   259 

Drawing  Room  Exercise  No.  5   295 

Drawing  Room  Exercise  No.  6   324 

Drawing  Room  Exercise  No.  7   413 

Drawing  Room  Exercise  No.  8   415 

Drawing  Room  Exercise  No.  9   582 

Economic  depth  of  plate  girders   180  to  181 

Economic  depth  of  trusses    575 

Economic  length  of  spans     ' 576 

Eccentrically  loaded  columns    112  to  114 

Elasticity,  definition  of    19 

Elastic    limit    19 

Equilibrium,  definition   of    31  to     35 

Equilibrium  of  couples     35 

Equilibrium  polygon    38 

Equilibrium  polygon  through  two  and  three  points   159  to  161 

Equation  of  the  elastic  curve    72  to     75 

Fabrication     7 

Fixed  beams   90  to     94 

Flange  stress  in  plate  girders    178  to  180 

Flange  rivets,  spacing 185  to  187 

Flange  splice    254 

Flange  increment 184  to  185 

Force,  definition    of    14 

Force,  measure  of   14 

Force,  direction  of  action    16 

Force,  line  of  action    '. 16 

Force  polygon 26 

Graphical  statics,  theoretical    144  to  161 

Graphical  determination  of  resultants  of  parallel  forces   35  to     43 

Graphical  determination  of  reactions  and   moments    147 

Graphical  analysis  of  deck  plate  girders   255  to  259 

Graphical  analysis  of  through   plate   girders    286  to  290 

Graphical  analysis  of  curved  chord  bridges    , 473  to  481 

Horizontal  shearing  stress  in  beams 65  to     68 

Highway  bridges    533  to  569 

Impact,  railroad  bridges    198 

Impact,  highway  bridges     533 

Increment  of  flange  stress    184  to  185 

Increment  of  bending  stress     64  to     65 

Inertia,  definition   of    47 

Inertia,  moment  of 48  to     50 

Influence  lines    162  to  171 

Length  of  cover  plates  (plate  girders)    181  to  184 

Lettering     9  to     10 

Live  load  for  railroad  bridges   195  to  197 

Live  load  for  highway   bridges    533  to  535 

Length  of  span    (economic)    576 

Manufacture  of  steel    1 

Maximum  shear  on  trusses 134  to  138 


INDEX  669 

PAGES 

Maximum  moment  on  trusses 138  to  139 

Mill  buildings,  design  of    583  to  624 

Modulus  of  elasticity 20 

Moment  of  a  force 23 

Moment  of  a  couple    23 

Motion    ^ 22 

Office  buildings,  design  of 624  to  652 

Parabola,  graphical  construction  of ,     61 

Pettit  trusses,  determination  of  stresses  in 481  to  502 

Pins,  determination  of  stress  in   118  to  120 

Plate  girders,  economic  depth    180  to  181 

Pony   trusses    (highway) 546  to  554 

Portals,  stresses  in   576  to  582 

Radius  of  gyration   48  to     50 

Railroad  bridges    195  to  533 

Rankine's  column  formula 106  to  107 

Reactions,  definition   of    17 

Reactions,  on  simple  beams    124 

Reactions,  on  simple  trusses     132 

Relation  of  stress  and   distortion    19  to     20 

Relation  of  shear  to  cross  bending 61  to     64 

Resultant  of  parallel  forces    43  to     44 

Rivets  and   riveting    3  to       5 

Rivets,  size  of   12 

Rivets,  conventional    signs    11 

Rivets,  stress  on   115  to  117' 

Rivets,  in  flanges  of  plate  girders 185  to  187 

Rollers,  theoretical  treatment 120  to  121 

Shafting,  theoretical  treatment    121  to  123 

Shear  on  beams    51  to     53 

Shear,  maximum  on  beams 124  to  128 

Shear,  maximum  simple  trusses     134  to  138 

Skew  bridges 570  to  571 

Specifications,  railroad    bridges    195 

Specifications,  highway   bridges    534  to  540 

Straight  line  column  formula   108  to  109 

Stiffeners     190  to  191 

Stress,  definition   of    17  to     19 

Stress,  maximum     68  to     72 

Stress  on  beams    51 

Stress  on  rivets     115  to  117 

Stress  on  trusses     139  to  143 

Stresses  in  curved  chord  bridges 415  to  426 

Structural  material     1  to       2 

Structural  shapes    2 

Structural  draughting    9  to     13 

Tracings 11 

Truss  bridges   (railroad) 324  to  532 

Types  of  railroad  bridges    195 

Types  of  bridge  trusses 324 

Viaducts,  railroad 295  to  324 

Warren  trusses 502  to  513 

Web   splice    189  to  190 

Whipple    trusses    514 

Wind  load,  railroad  bridges    198 

Wind  load,  highway  bridges   535 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

AN     INITIAL     FINE     OF    25     CENTS 

WILL   BE   ASSESSED    FOR    FAILURE  TO    RETURN 

HIS   BOOK   ON   THE   DATE  DUE.     THE   PENALTY 

WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 

DAY    AND    TO     $1.OO    ON     THE    SEVENTH     DAY 


MAR    4 
MAR  19  1934 


FEB  141936 


MAY     7  1937 


OCT  291940 


1943 


OCT   3  1934 
APR  271935 


LD  21-50m-l,'3l 


Y.C  40403 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


